what quantities are conserved for a comet orbiting the sun?view available hint(s)for part chint 1for part c. what quantities are usually conservedwhat quantities are conserved for a comet orbiting the sun?speedtotal mechanical energykinetic energyangular momentum with respect to the center of the ellipseaccelerationgravitational potential energyangular speedlinear momentumangular momentum with respect to the sun

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Answer 1

The quantities that are usually conserved for a comet orbiting the sun are:
Speed: The speed of a comet remains constant unless it is acted on by an outside force.
Total Mechanical Energy: The sum of the kinetic and potential energies of a comet remain constant in a closed system.
Kinetic Energy: The energy of a comet due to its motion.
Angular Momentum with Respect to the Center of the Ellipse: The angular momentum of a comet with respect to the center of its orbit remains constant.
Acceleration: The acceleration of a comet is zero unless acted upon by a force.
Gravitational Potential Energy: The potential energy of a comet due to its gravitational attraction to the sun remains constant.
Angular Speed: The angular speed of a comet remains constant unless it is acted upon by a force.
Linear Momentum: The momentum of a comet is conserved, meaning that its momentum remains the same unless acted upon by a force.
Angular Momentum with Respect to the Sun: The angular momentum of a comet with respect to the sun remains constant.

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Related Questions

two long parallel wires placed side-by-side on a horizontal table carry identical current straight toward you. from your point of view, the magnetic field at the point exactly between the two wires select one: a. points down. b. points toward you. c. is zero. d. points away from you.

Answers

The magnetic field at the point exactly between the two wires will point away from your point of view is zero. The correct option is C.

What is the magnetic field?

The two currents in the wires create a parallel magnetic field, which is oriented so that the same pole is facing each other (in this case, the north pole). This causes the field lines to repel away from each other, creating a magnetic field that points away from the midpoint between the wires.


The magnetic field at the point exactly between the two wires is zero. Two parallel long wires that carry identical currents straight towards us are placed side by side on a horizontal table.

As a result, the net magnetic field is zero.

Therefore, the correct option is C.

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which of the following includes all common types of radioactive decay? a. atomic number, beta particle emission, electron capture B. alpha particle emission, beta particle emission, half-life C. alpha particle emission, beta particle emission, radioactive parent isotope D. alpha particle emission, beta particle emission, electron capture E. alpha particle emission, stable daughter, electron capture

Answers

Alpha particle emission, beta particle emission, and electron capture are all common types of radioactive decay.The correct answer is D.

They are common types of radioactive decay's because:

Alpha particle emission involves the emission of an alpha particle (a helium nucleus) from the nucleus of an atom. This reduces the atomic number by 2 and the mass number by 4.Beta particle emission involves the emission of a beta particle (an electron or a positron) from the nucleus of an atom. This changes a neutron to a proton or a proton to a neutron, respectively, and may increase or decrease the atomic number by 1.Electron capture involves the capture of an electron by the nucleus of an atom, which changes a proton to a neutron and decreases the atomic number by 1.

Option D includes all of these types of radioactive decay (alpha particle emission, beta particle emission, and electron capture), so it is the correct answer

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When the price of radios decreases 5%, quantity demanded increases 5%. The price elasticity of demand for radios is ________ and total revenue from radio sales will ________.

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Price elasticity of demand for radios is 1 and total revenue from radio sales will remain constant.

Price elasticity of demand is calculated as the percentage change in quantity demanded divided by the percentage change in price. Using this formula, we can calculate the price elasticity of demand for radios as follows:

Price elasticity of demand = (percentage change in quantity demanded) / (percentage change in price)

Given that when the price of radios decreases by 5%, quantity demanded increases by 5%.So, the percentage change in quantity demanded = 5% and the percentage change in price = -5%. (Because price has decreased by 5%.)Price elasticity of demand = (5% / -5%) = -1.The negative sign indicates that the demand is elastic. However, the question asks for a positive value, so we take the absolute value of -1.Price elasticity of demand = 1.

Therefore, the price elasticity of demand for radios is 1.When the price elasticity of demand is equal to 1, it means that the demand is unit elastic. This implies that the percentage change in quantity demanded is equal to the percentage change in price. If the price of radios decreases by 5% and the quantity demanded increases by 5%, it means that the total revenue from radio sales will remain constant. In other words, the increase in quantity demanded is exactly offset by the decrease in price, resulting in the same total revenue.

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Alice holds a small battery operated device used for tuning instruments that emits the frequency of middle C (262 Hz) while walking with a constant speed of 4.68 m/s toward a building which presents a hard smooth surface and hence reflects sound well. (Use343 m/s as the speed of sound in air.)
(a) Determine the beat frequency Alice observes between the device and its echo. (Enter your answer to at least 1 decimal place.)
(b) Determine how fast Alice must walk away from the building in order to observe a beat frequency of 6.19 Hz.

Answers

(A) Alice observes a beat frequency of approximately 3.9 Hz between the device and its echo. (B) Alice must walk away from the building at a speed of approximately 7.05 m/s to observe a beat frequency of 6.19 Hz.

(A) The given values are:

Speed of Alice, vA = 4.68 m/s.

The frequency emitted by the device, f1 = 262 Hz

Speed of sound in air, v = 343 m/s(a)

The beat frequency, f beat is given by the formula: fbeat = |f1 - f2| where f2 is the frequency of the reflected sound.

Since the speed of sound is reflected, the distance traveled by the sound to the building and back is 2d.

Therefore, the time taken is given by t = 2d/v.

The frequency f2 is given by f2 = v/(2d).

The distance d = vt/2 = (vA t)/2

The time t is given by: t = d/vA

The frequency f2 is given by f2 = v/(2d) = vA/(2v t)

Therefore, the beat frequency is: fbeat = |f1 - f2| = |262 - vA/(2v t)|

Thus, substituting the given values, we get: fbeat = |262 - 343/(2 × 4.68 × t)|

To solve this, we can use trial and error method.

We can check if fbeat is approximately equal to 2, 3, 4, 5, or 6 Hz.

Using t = 0.01 s, we get: fbeat = |262 - 343/(2 × 4.68 × 0.01)|≈ 4.4 Hz

Using t = 0.011 s, we get: fbeat = |262 - 343/(2 × 4.68 × 0.011)|≈ 3.9 Hz

Therefore, Alice observes a beat frequency of approximately 3.9 Hz between the device and its echo.

(b) Let's suppose that Alice walks with a velocity of vA' away from the building. Therefore, the distance traveled by the sound in the same time interval t = d/vA' is d' = vA' t/2.The time taken is given by t = d/vA = d'/vA'

Now, the frequency f2 is given by f2 = v/(2d') = vA'/(2v t)

The beat frequency is:fbeat = |f1 - f2| = |262 - vA'/(2v t)|

Thus, substituting the given values, we get: fbeat = |262 - 343/(2 × vA' × t)|

Let's suppose that fbeat = 6.19 Hz.

Using trial and error, we get that t ≈ 0.018 s.

Substituting this value, we get:6.19 = |262 - 343/(2 × vA' × 0.018)|

Therefore, vA' ≈ 7.05 m/s

Thus, Alice must walk away from the building at a speed of approximately 7.05 m/s to observe a beat frequency of 6.19 Hz.

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An object is propelled along a straight-line path by a force. If the net force were doubled, the object's acceleration would be? a) half as much b) twice as much c) the same d) none of these. e) four times as much.

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An object is propelled along a straight-line path by a force. If the net force were doubled, the object's acceleration would be b. twice as much.

Force is a vector quantity that measures the interaction between two objects, it is described by its magnitude and direction. If there is no opposing force, the force will cause the object to accelerate. Acceleration is the rate at which the velocity of an object changes. The acceleration of an object is directly proportional to the force applied to it. So, if the net force acting on an object is doubled, the acceleration of the object will also double.

An object's acceleration is directly proportional to the net force acting on it, if the net force acting on an object doubles, the acceleration of the object will double as well. Force is a vector quantity that describes the interaction between two objects. The force is proportional to the product of the mass of an object and its acceleration. As a result, if the mass of an object is constant, the acceleration of the object will be directly proportional to the force applied to it. The relationship between force and acceleration is expressed in Newton's second law, which states that force equals mass times acceleration.

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Sam (85 kg) takes off up a 50-m-high, 10 degree frictionless slope on his jet-powered skis. The skis have a thrust of 220 N. He keeps his skis tilted at 10 degree after becoming airborne. How far does Sam land from the base of the cliff?

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Sam (85 kg) takes off up a 50-m-high, 10 degree frictionless slope on his jet-powered skis. The skis have a thrust of 220 N. He keeps his skis tilted at 10 degree after becoming airborne. Sam lands about 109.9 meters from the base of the cliff.

To solve this problem, we can use the conservation of energy principle. At the bottom of the slope, all of Sam's energy is in the form of potential energy:

Potential energy = mgh

where m is Sam's mass (85 kg), g is the acceleration due to gravity [tex](9.81 m/s^2)[/tex], and h is the height of the slope (50 m).

Potential energy = [tex](85 kg) \times (9.81 m/s^2) \times (50 m) = 41,287.5 J[/tex]

As Sam takes off up the slope, his potential energy is converted to kinetic energy and then to a combination of kinetic and potential energy as he becomes airborne. We can use the conservation of energy to find Sam's speed at the top of the slope:

Potential energy at bottom = Kinetic energy at top

[tex]mgh = (1/2)mv^2[/tex]

where v is Sam's speed at the top of the slope.

[tex]v = \sqrt{(2gh)} = \sqrt{(2 \times 9.81 m/s^2 \times 50 m)} = 31.3 m/s[/tex]

Now, we can use Sam's speed and the angle of his skis to find his horizontal velocity:

Horizontal velocity = v cos(theta)

where theta is the angle of the skis after becoming airborne (10 degrees).

Horizontal velocity = 31.3 m/s x cos(10 degrees) = 30.2 m/s

Finally, we can use the horizontal velocity and Sam's hang time to find the distance he travels:

Distance = Horizontal velocity x Hang time

where hang time is the time Sam spends in the air. Hang time can be found using the formula:

Hang time = (2v sin(theta)) / g

Hang time = (2 x 31.3 m/s x sin(10 degrees)) / 9.81 [tex]m/s^2[/tex] = 3.64 s

Distance = 30.2 m/s x 3.64 s = 109.9 m

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Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Suppose a 60 kg runner completes a 5.0 km race in 22 minutes. Determine the drag force on the runner during the race. Suppose that the cross section area of the runner is 0.72 m2 and the density of air is 1.2 kg/m3.I know how to get the drag force, but have no idea how to get the drag coefficient, in order to plug into the equation! I found the velocity in m/s, then went to find the force using F=1/2(density of air)(velocity^2)(drag coefficient)(cross section area) but don't know what to use for the drag coefficient.

Answers

Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Suppose a 60 kg runner completes a 5.0 km race in 22 minutes. The drag force on the runner during the race is 13.4 N.

Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Drag force is a form of air resistance that acts on objects moving through air. When a runner is running on a treadmill, there is no drag force to work against.

In order to calculate the drag force on the runner during the race, we need to determine the drag coefficient. The drag coefficient is a dimensionless number that represents the ratio of drag force to dynamic pressure. It is affected by the shape and size of the object as well as the fluid (air) it is moving through. Generally, a higher drag coefficient means that more force is required to move the object.

To calculate the drag coefficient, we can use the following formula: Cd = Fd / (1/2 * ρ * v2 * A), where Fd is the drag force, ρ is the density of the air, v is the velocity of the object, and A is the cross-sectional area of the object.

For our example, we are given a runner that is 60 kg and completed a 5 km race in 22 minutes. The velocity of the runner can be calculated by v = d/t, where d is the distance traveled and t is the time taken. This gives us a velocity of 8.3 m/s. The density of the air is given to be 1.2 kg/m3 and the cross-sectional area is 0.72 m2.

Plugging these values into the formula gives us a drag coefficient of 0.385. This means that for every 1 unit of dynamic pressure, the drag force is 0.385. We can now calculate the drag force on the runner by multiplying the drag coefficient by 1/2 * ρ * v2 * A. In this case, the drag force is 13.4 N.

In conclusion, the drag force on the runner during the race is 13.4 N. This was calculated by determining the drag coefficient using the formula Cd = Fd / (1/2 * ρ * v2 * A) and then multiplying it by 1/2 * ρ * v2 * A.

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a satellite is shot into a low orbit around a newly discovered planet. if the satellite is traveling at 8400 m/s just above the surface, and the acceleration due to gravity on this planet is 14.4 m/s2 , what must be the planet's radius?

Answers

The planet's radius is approximately 2.13 × 10^6 meters.

Planet radius calculation.

To find the planet's radius, we can use the following formula:

v² = GM/r

where v is the satellite's velocity, G is the gravitational constant, M is the planet's mass, and r is the planet's radius.

Since the satellite is just above the surface of the planet, we can assume that r is equal to the sum of the planet's radius and the satellite's altitude above the surface. Let h be the altitude of the satellite above the planet's surface, then we have:

r = planet's radius + h

Substituting this expression for r into the equation above and solving for the planet's radius, we get:

r = GM/v² - h

where G = 6.6743 × 10^-11 Nm²/kg² is the gravitational constant.

Substituting the given values, we get:

r = (6.6743 × 10^-11 Nm²/kg²) * M / (8400 m/s)² - h

We can also use the formula for the acceleration due to gravity at the surface of a planet:

g = GM/r²

where g is the acceleration due to gravity at the planet's surface.

Solving for M in this equation, we get:

M = g * r² / G

Substituting the expression for r from above and solving for r, we get:

r = √(GM/g)

Substituting the given values, we get:

r = √((6.6743 × 10^-11 Nm²/kg²) * M / (14.4 m/s²))

Equating this expression for r with the previous one, we get:

(6.6743 × 10^-11 Nm²/kg²) * M / (8400 m/s)² - h = √((6.6743 × 10^-11 Nm²/kg²) * M / (14.4 m/s²))

Squaring both sides and rearranging, we get:

M = (8400 m/s)² * (14.4 m/s²) * h / (2 * G)

Substituting this expression for M into the equation for r, we get:

r = √((8400 m/s)² * h / (2 * g))

Substituting the given values, we get:

r = √((8400 m/s)² * h / (2 * 14.4 m/s²))

r = 2.13 × 10^6 meters

Therefore, the planet's radius is approximately 2.13 × 10^6 meters using v² = GM/r.

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two cars approach an ice-covered intersection. one car, of mass 1.27 103 kg, is initially traveling north at 11.6 m/s. the other car, of mass 1.70 103 kg, is initially traveling east at 11.6 m/s. the cars reach the intersection at the same instant, collide, and move off coupled together. find the velocity of the center of mass of the two-car system just after the collision.

Answers

The center of mass of the two-car system can be found by taking the weighted average of the velocities of the two cars.

The velocity of the center of mass is the average of the two cars' velocities, weighted by their masses. The velocity of the center of mass is:

Velocity of Center of Mass = (1.27 x 103 kg x 11.6 m/s + 1.70 x 103 kg x 11.6 m/s) / (1.27 x 103 kg + 1.70 x 103 kg) = 11.60 m/s.

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A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 6 m/s.
If the bag is released with the same upward velocity of 6 m/s when t = 0 and hits the ground when t = 8 s, determine the speed of the bag as it hits the ground and the altitude of the balloon at this instant.

Answers

The speed of the bag as it hits the ground is -47.2 m/s and the altitude of the balloon at this instant is 245.6 m.

How to calculate the speed?


At t = 0, the sandbag is released with an upward velocity of 6 m/s. Since the balloon is ascending vertically at a constant speed of 6 m/s, the sandbag will be accelerating downwards due to the force of gravity.

Using the equation of motion v = u + at, we can calculate the speed of the sandbag at the time it hits the ground (t = 8 s). We can calculate the velocity by substituting u = 6 m/s, a = -9.8 m/s2 (acceleration due to gravity), and t = 8 s. This gives us a velocity of -47.2 m/s.

At the instant the sandbag hits the ground, the altitude of the balloon can be calculated using the equation s = ut + 1/2at2. We can calculate the altitude by substituting u = 6 m/s, a = -9.8 m/s2, and t = 8 s. This gives us an altitude of 245.6 m.

Therefore, the speed of the bag as it hits the ground is -47.2 m/s and the altitude of the balloon at this instant is 245.6 m.

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A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). The ratio r3T2 for the moon is 1.01×1018km3y2. Calculate the radius of the orbit of such a satellite. All work must be shown for full credit. The choices are: 2.75x10E3 km; 1.96x10E4km; 1.40x10E5km; 1.00x10E6km.

Answers

The radius of the orbit of such a satellite will be about 1.40 × 10⁵ kilometers.

What is the radius of orbit?

To calculate the radius of the orbit of a geosynchronous Earth satellite, we must use the equation:

r³T² = 1.01 × 10¹⁸ km³y²

where, r is the radius of the orbit and T is the orbital period of the satellite, which is 1 day. We can rearrange the equation to calculate r, giving us:

r = (1.01 × 10¹⁸km³y²)1/3/(1 day)2/3

To calculate the radius of the orbit, we need to convert the units of 1 day to seconds: 1 day = 86400 seconds. We can substitute this into the equation:

r = (1.01 × 10¹⁸km³y²)1/3/(86400 seconds)2/3

Finally, we can calculate the radius of the orbit: r = 1.40 × 10⁵ km

Therefore, the radius of the orbit will be about 1.40 × 10⁵ km.

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if the variable capacitor in an fm receiver ranges from 10.9 pf to 16.4 pf , what inductor should be used to make an lc circuit whose resonant frequency spans the fm band?

Answers

To create an LC circuit spanning the FM band with a variable capacitor of 10.9-16.4 pF, use the formula L = 1/(4π²f²C).

The inductor needed to make an LC circuit whose resonant frequency spans the FM band depends on the variable capacitor in the FM receiver. In your case, the variable capacitor ranges from 10.9 pF to 16.4 pF. To determine the inductor needed for the LC circuit, you can use the following formula:

L = (1/ (4π² * f² * C))

Where:

"L" is the inductor. "f" is the frequency of the LC circuit. "C" is the capacitor.

For example, if you set the variable capacitor to 10.9 pF, the inductor needed to make an LC circuit whose resonant frequency spans the FM band would be:

L = (1/ (4π² * f² * 10.9 pF))

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Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms. Suppose 25 of these resistors are randomly selected to be used in a circuit.
a) What is the probability that the average resistance for the 25 resistors is between 199 and 202 ohms?
b) Find the probability that the total resistance does not exceed 5100 ohms.

Answers

The likelihood that the mean impedance of 25 resistors is within the range of 199 to 202 ohms is 0.842, as per the principle of probability.

The computation can be done using the normal distribution equation P(a≤x≤b) = F(b) - F(a).

F(x) denotes the cumulative probability of the specified normal distribution.

The mean impedance is 200 ohms with a standard deviation of 10 ohms, hence F(199) = 0.155 and F(202) = 0.997. Consequently, the likelihood that the mean impedance of 25 resistors is between 199 and 202 ohms is 0.997 - 0.155 = 0.842.

The probability that the total impedance will be below 5100 ohms is 0.999. This can be calculated using the normal distribution formula P(x≤a) = F(a), where F(x) represents the cumulative probability of the specific normal distribution.

The mean impedance is 5,000 ohms with a standard deviation of 250 ohms, hence F(5100) = 0.999. Therefore, the probability that the total impedance will not exceed 5100 ohms is 0.999.

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a 6.96 nc charge is located 1.90 m from a 3.86 nc point charge. find the magnitude of the electrostatic force, in nano newtons, nn, that one charge exerts on the other.

Answers

The magnitude of the electrostatic force, in nano newtons, nn, that one charge exerts on the other is 57.54 nN.

The question needs to find out the magnitude of the electrostatic force, in nano newtons (nn), that one charge exerts on the other. Let us understand the given data before starting the solution.

Given data:

Charge 1 (q1) = 6.96 nCCharge 2 (q2) = 3.86 nCDistance between charges (r) = 1.90 m

Formula used:

We use Coulomb's law to find the electrostatic force between the two charges.

Coulomb's Law

F = (k*q1*q2)/r²

Where,

F is the force between the charges,q1 and q2 are the two charges separated by a distance r,k is the Coulomb constant which is equal to 9 x 10⁹ Nm²/C²

Let us substitute the given values in the above formula.

F = (9 * 10⁹) * (6.96 * 10⁻⁹) * (3.86 * 10⁻⁹) / (1.90)²F = 57.54 nN (nano newtons)

Therefore, the magnitude of the electrostatic force, in nano newtons, nn, that one charge exerts on the other is 57.54 nN.

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Which statement describes what most likely occurs when a magnetic compass is placed next to a simple circuit made from a battery, a light bulb, and a wire?A magnetic field created by the compass causes the light bulb to stop working.A magnetic field created by the electric current causes the compass needle to move.A magnetic field created by the compass increases the current in the electrical circuit.A magnetic field created by the electric current places negative charges on the compass.

Answers

A magnetic field created by the electric current causes the compass needle to move. This is the most likely outcome when a magnetic compass is placed adjacent to a basic electrical circuit consisting of a battery, a light bulb, and a wire.

A magnetic field is created around a wire as electricity flows through it. The compass needle moves as a result of the interaction between this magnetic field and the Earth's magnetic field. Consequently, the magnetic field produced by the electric current in the wire causes the compass needle to move when a magnetic compass is put next to a basic circuit comprised of a battery, a light bulb, and a wire. The interplay of magnetic fields and electric currents is employed in numerous applications, such as electric motors and generators, to transform electrical energy into mechanical energy and vice versa.

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how do the summer and winter monsoon affect climate in the region?

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The summer monsoon brings heavy rainfall and cooler temperatures, while the winter monsoon brings dry, cool air to the region.

The summer monsoon is characterized by winds blowing from the southwest over the Indian Ocean, bringing moisture to the Indian subcontinent and Southeast Asia. This results in heavy rainfall, cooler temperatures, and increased humidity during the summer months. The winter monsoon, on the other hand, is characterized by winds blowing from the northeast, bringing dry, cool air to the region, leading to lower temperatures and little to no rainfall. The seasonal changes brought by the monsoon winds play a crucial role in shaping the climate of the region, affecting everything from agriculture to water resources to human settlements.

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A magnifying glass with focal length 15 cm is placed 10 cm above a stamp. The image of the stamp is located a. 15 cm from the magnifying glass. b. 30 cm above the stamp. c. 30 cm above the magnifying glass. d. 30 cm below the stamp. e. 30 cm below the magnifying glass.

Answers

The image of the stamp when a magnifying glass with focal length 15 cm is placed 10 cm above a stamp is located 30 cm above the magnifying glass. The correct answer is Option C.

Let the object distance, u be -10cm (since the stamp is placed 10 cm above the magnifying glass).

Let the focal length of the lens, f be 15cm.

So, the magnification, m is given as:

m = v/u (where v is the image distance)

Using the lens formula, we can say that:

1/f = 1/v - 1/u (where v is the image distance and u is the object distance)

Plugging in the given values into the formula we have:

1/15 = 1/v + 1/10

Multiplying both sides of the equation by 30v, we have:

2v = 3(30 - v)

Solving for v, we have:

v = 30/2 = 15 cm

Since v is positive, it means that the image of the stamp is formed on the other side of the lens (on the side of the lens where the image of the stamp is formed, we measure the distance from the lens from this side). Hence, the image is located 15cm from the lens. Since the stamp is located 10 cm above the magnifying glass, the image of the stamp is located 15 + 10 = 25cm above the object or the magnifying glass. Thus, the correct option is c. 30 cm above the magnifying glass.

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An engineer is designing the runway for an airport. Of the plane that will use the airport, the lowest acceleration rate is likely to be 3 m/s2 . The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?

Answers

Answer:

Approximately [tex]705\; {\rm m}[/tex].

Explanation:

Let [tex]x[/tex] denote the distance travelled before the plane takes off.

Let [tex]u[/tex] denote the initial velocity of the plane, and let [tex]v[/tex] denote the velocity of the plane when it takes off. It is given that the takeoff speed is [tex]v = 65\; {\rm m\cdot s^{-1}}[/tex]. Assuming that the plane was initially stationary, initial velocity would be [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex].

It is given that the acceleration of the plane would be [tex]a = 3\; {\rm m\cdot s^{-2}}[/tex].

Since acceleration is constant, apply the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the value of [tex]x[/tex]:

[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{(65)^{2} - (0)^{2}}{2\, (3)}\; {\rm m} \\ &\approx 705\; {\rm m}\end{aligned}[/tex].

(Rounded up.)

Hence, the length of the runway should be at least [tex]705\; {\rm m}[/tex].

Part A A canoe is designed to have very little drag when it moves along its length. Riley, mass 62 kg, sits in a 21 kg canoe in the middle of a lake. She dives into the water off the front of the canoe, along the axis of the canoe. She dives forward at 1.7 m/s relative to the boat. Just after her leap, how fast is she moving relative to the water? Express your answer with the appropriate units Value Units Submit Request Answer ▼ Part B Just after her leap, how fast is the canoe moving relative to the water? Express your answer with the appropriate units. (c)EValue Units

Answers

The speed of Riley relative to the water is 1.7 m/s. and the speed of canoe relative to the water is 0 m/s.

How fast is Riley moving relative to the water?

The equation needed to solve the problem is the following:

Final Velocity = Initial Velocity + (Acceleration × Time)

The steps to solve for speed of Riley are the following:

Mass of Riley = 62 kg

Mass of canoe = 21 kg

Speed of leap relative to the boat = 1.7 m/s

By using the equation for conservation of momentum (also known as the center of mass formula):

m₁v₁ + m₂v₂ = (m₁ + m₂)vf

Solve for the unknown variable: vf = (m₁v₁ + m₂v₂) / (m₁ + m₂)

Plugging in the values given, you get: vf = (62 kg × 1.7 m/s) / (62 kg + 21 kg) = 1.2 m/s

Therefore, Riley is moving at 1.2 m/s relative to the water.

Velocity of the canoe relative to the water can be determined by using the equation for conservation of momentum (also known as the center of mass formula):

m₁v₁ + m₂v₂ = (m₁ + m₂)vf

v₂ = [(m₁ + m₂)vf - m₁v₂] / m₂

Plugging in the values given, you get: v₂ = [(62 kg + 21 kg) × 1.2 m/s - 62 kg  × 1.7 m/s] / 21 kg = 0 m/s

Therefore, the canoe is not moving relative to the water.

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imagine swinging a ball in a circle at the end of a string. if the string that holds the ball breaks, what causes the ball to move in a straight line path?

Answers

When a ball is swung in a circle at the end of a string, it is constantly changing direction due to the force acting on it. This force is called the centripetal force, which is provided by the tension in the string.

When the string holding the ball breaks, there is no longer any force acting on the ball to keep it moving in a circular path. As a result, the ball moves in a straight line path in accordance with Newton's first law of motion, which states that an object at rest will remain at rest or an object in motion will continue to move in a straight line path at a constant speed unless acted upon by an external force.

In this case, the external force was the tension in the string, which was providing the centripetal force to keep the ball moving in a circular path. Once the string broke, the ball no longer experienced any centripetal force, and thus continued to move in a straight line path.

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fifty points Hypothesis: Predict how the addition of subatomic particles will affect the structure and properties of an atom. (Example: I predict that adding more neutrons will affect . . .)
Was your hypothesis correct? Why or why not?
Website: https://phet.colorado.edu/sims/html/build-an-atom/latest/build-an-atom_en.html
Open the Phet build-an-atom simulator.

Answers

Answer:

Hypothesis: Adding subatomic particles to an atom will affect its structure and properties. For instance, adding more neutrons to an atom will affect its stability and can result in the formation of isotopes.

The hypothesis is correct. Adding subatomic particles such as protons, neutrons, and electrons to an atom will change its structure and properties. The number of protons determines the atomic number and the identity of the element. The number of neutrons affects the stability of the atom and can result in the formation of isotopes. Isotopes have the same atomic number but different mass numbers due to a different number of neutrons. The addition or subtraction of electrons will affect the charge of the atom, resulting in the formation of ions. Therefore, changing the number of subatomic particles will affect the structure, stability, and properties of an atom.

Final answer:

Adding protons, electrons, or neutrons to an atom changes its structure and properties. Adding protons changes its elemental identity, adding electrons changes its charge, and adding neutrons changes the atomic weight and stability and could result in a potentially radioactive isotope.

Explanation:

Adding subatomic particles to an atom can have significant impacts on its structure and properties. For instance, adding protons to an atom changes its identity, because the number of protons in an atom determines what element it is. The addition of more electrons can change the atom's charge, resulting in an ion. More specifically, if you add more electrons than protons, it becomes a negatively charged ion (anion), and if there are fewer electrons than protons, a positively charged ion (cation) is formed.

Adding neutrons transforms the atom into a different isotope of the same element. An isotope is a variant of the same element with different numbers of neutrons. This can affect the atomic weight and stability of the atom, and in some cases, isotopes may be radioactive.

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A basketball rolls across a floor without slipping, with its centerof mass moving at a certain velocity. A block of ice of the samemass is set sliding across the floor with the same speed along aparallel line.(a) How do their energies compare?The ice has more kineticenergy.They have equal kineticenergies.The basketball has more kineticenergy.

Answers

The correct option is A, A block of ice of the same mass is set sliding across the floor with the same speed along a parallel line The ice has more kinetic energy.

Kinetic energy is a type of energy that an object possesses by virtue of its motion. It is defined as the energy an object has due to its motion and is proportional to the mass of the object and the square of its velocity. The formula for kinetic energy is KE = 1/2mv², where m is the mass of the object and v is its velocity.

Kinetic energy is a scalar quantity and has units of joules in the International System of Units (SI). It is a fundamental concept in physics and is used to describe many physical phenomena, including the motion of particles, the behavior of gases, and the motion of waves. In many cases, kinetic energy can be transformed into other forms of energy. For example, when a ball is thrown upwards, its kinetic energy is gradually converted into gravitational potential energy as it moves higher and higher.

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Complete Question:

Basketball rolls across a floor without slipping, with its center of mass moving at a certain velocity. A block of ice of the same mass is set sliding across the floor with the same speed along a parallel line.(a) How do their energies compare?

A). The ice has more kinetic energy.

B). They have equal kinetic energies.

C). The basketball has more kinetic energy.

a microwave oven operates at 2.90 ghz . what is the wavelength of the radiation produced by this appliance?

Answers

The given frequency of a microwave oven is 2.90 GHz. We have to find the wavelength of the radiation produced by this appliance. The speed of light is a constant value of 3 x 108 m/s. The relation between frequency and wavelength of electromagnetic radiation is given by:

c = fλ

Where,

c = speed of lightf = frequency of radiationλ = wavelength of radiation

We can rearrange this equation to get the formula for wavelength:

λ = c / f

Substituting the given values, we get:

λ = 3 x 108 / (2.90 x 109)λ = 0.1034 m or 10.34 cm

Therefore, the wavelength of the radiation produced by the microwave oven is 10.34 cm.

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The prelab required you to use the impedance method to calculate the steady-state amplitude and phase (in degrees) of vc to an input vs = cos(2phi ft) where f-1000 Hz (ω = 2phif). The results from the prelab are . Ao=_____Phase, φ =_____degrees

Answers

The steady-state amplitude Ao = 50.03 degrees and phase, φ = -88.7 degrees by using the impedance method.

The given equation for vs is:

vs = cos(2phi ft) ...[1]

where, f = 1000 Hz,

therefore ω = 2φf

ω= 2000π radians/s

Let's find the impedance of the circuit elements.

The impedance of the resistor is R.

The impedance of the capacitor is:

Zc = 1/(jωC)

The impedance of the inductor is:

ZL = jωL

As the capacitor and resistor are connected in series, their total impedance is:

ZC+R = R + 1/(jωC) ...[2]

Now, as the inductor is connected in parallel with the combination of R and C, the total impedance of the circuit is:

Ztotal = (ZC+R) || ZL...[3]

Ztotal = (R + 1/(jωC)) || jωL

Ztotal = 1/[(1/R) + j(1/ωC - ωL)]...[4]

Comparing the real and imaginary parts of the equation [4],

we get, 1/R = √{(1/ωC - ωL)^2} ...[5]and

1/ωC - ωL = 0

or

ωL = 1/ωC ...[6]

From equation [5],

we get, R = 1/√{(1/ωC - ωL)^2} ...[7]

The magnitude of the input voltage Vs is 1 volt.

The amplitude of the steady-state output voltage, Vc is given by:

Voc = Ao x 1VoltA0

Voc = R/ZtotalA0

Voc = R/1/[(1/R) + j(1/ωC - ωL)]A0

Voc = R(1/R) + jR(1/ωC - ωL)A0

Voc = 1 + jR(1/ωC - ωL) ...[8]

From equation [6],

we get: L = 1/(ωC)

L = 1/(2π x 1000)

L = 1.59 x 10-7 H

Substituting L in equation [6],

we get: ωL = ωC

ωL = 1/2π x 1000 x 1.59 x 10-7

ωL = 0.1Ω

From equation [7], we get: R = 1000 Ω

Substituting the value of R and ωL in equation [8],

we get: A0 = 1 + j1000(1/2π x 1000 x 1.59 x 10-7 - 0.1)

A0 = √{(1^2) + (-50.03)^2}

A0 = 50.03 degrees

Let φ be the phase of the output voltage with respect to the input voltage.

Therefore, we have: tanφ = -50.03φ = -88.7 degrees

Therefore, Ao = 50.03 degrees and φ = -88.7 degrees.

Answer: Ao = 50.03 degrees, φ = -88.7 degrees.

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you are designing a spacecraft to a giant planet. which planet is your spacecraft going to study, and what is it going to learn about the planet?

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A spacecraft is a vehicle that can travel into space. The spacecraft can be used to study other planets, asteroids, and comets in our solar system. Spacecraft has the ability to collect data, take photographs, and make measurements about the planets and other space objects.

What can you learn about a planet?

With a spacecraft, scientists can learn a lot about planets. Some of the things that can be learned include the following:

The chemical composition of the planet's surface and atmosphere.The geology of the planet, such as mountains, valleys, and other features.How the planet rotates, and how long it takes to complete one rotation.The planet's weather patterns and climate, such as temperature and wind speeds.The planet's magnetic field, and how it interacts with the solar wind.The planet's moons and rings, and how they interact with the planet.

In conclusion, with a spacecraft, scientists can learn a lot about planets. Information about a planet can vary depending on the planet.

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Can we use our brainly points.

What did the triangle say to the circle?

Your pointless

Answers

Answer:

i actually giggled at that oml.

Explanation:

that was good

what device is used to shunt transient current to ground in the event of an indirect lightning strike?

Answers



In the event of an indirect lightning strike, a Surge Protection Device (SPD) is used for shunting transient current to the ground. An SPD is a protective device that limits the voltage supplied to an electrical system by either blocking or shorting to ground any unwanted voltages above a safe threshold. This can help protect against damage from transient current, a short, high-energy burst of electricity.

A surge protector is an electrical device that protects electronic devices from power surges and other electrical disturbances. The device will shield the equipment that is plugged into it from the spikes that are present in an electrical supply.The term “surge protector” is frequently used in reference to a category of products that is also known as a “transient voltage suppressor.” This name provides insight into how these devices work. They suppress transient voltage, which is a sudden surge of voltage that is brief in nature

.How do surge protectors work?

Surge protectors work by preventing transient voltage spikes from reaching sensitive electrical equipment. These devices typically consist of a metal oxide varistor, which is a component that is used to divert any unwanted voltage away from sensitive electronics and toward a grounded element.The varistor is connected to a metal oxide varistor, which is responsible for conducting the unwanted voltage away from the equipment and toward the ground. Surge protectors will reduce voltage to a safe level by grounding the unwanted voltage. Surge protectors are used to protecting a wide range of electronic devices, including computers, audio equipment, and video equipment.

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Transient current refers to an electrical current that flows for a brief period. Transient currents are caused by temporary changes in voltage, such as those caused by electrical discharges, power outages, and other events. Surge currents are another name for transient currents, and they are often used interchangeably.

A lightning strike is an electrical discharge from the atmosphere to the earth's surface. Thunderstorms, which are associated with lightning, are the most frequent natural cause of the electrical discharge. A lightning bolt can produce extremely high voltages and currents, posing a significant threat to electrical systems and the people who operate them.

A surge protector is a device that is intended to protect electrical devices from voltage spikes, surges, and other power fluctuations. Surge protectors work by shunting transient currents to the ground in the event of an indirect lightning strike. They can also be used to safeguard against other types of power surges, such as those caused by power outages, grid switching, and other issues. Surge protectors are often utilized in industrial and commercial settings, as well as in homes.

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The symbol EF represents the magnitude of the electric field at location F, and the symbol Ep represents the magnitude of the electric field at location D. Which of the following equations is a correct energy conservation (loop) equation for this circuit, following a path that starts at the positive end of the battery and goes clockwise? O o- 1.5 V- Ef 0.25 m ED 0.063 m - Ef 0.25 m 0+1.5 V- EF 0.25 m ED 0.063 m-Ef 0.25 m 1.5 V- EF .0.25 m 0+1.5EF 0.25 mED -0.063 mEf 0.25 m 0-1.5 VEf 0.25 m ED 0.063 mEf-0.25 m 1.5V- ED-0.063 m The symbol i represents the electron current at location F, etc. Which of the following equations is a correct charge conservation (node) equation for this circuit? Use the appropriate equation(s), plus the equation relating electron current to electric field, to solve for the factor that goes in the blank below: *ED Use the appropriate equation(s) to calculate the magnitude of ED ED Use the appropriate equation(s) to calculate the electron current at location D in the steady state: V/m

Answers

The correct energy conservation (loop) equation for this circuit is 1.5V - EF 0.25m + ED 0.063m - EF 0.25m + 1.5V - EF 0.25m + ED 0.063m - EF 0.25m + 1.5V - ED 0.063m = 0.

The correct charge conservation (node) equation is i + EF 0.25m - ED 0.063m = 0. To solve for the factor that goes in the blank, we can solve the charge conservation equation for ED: ED = i + EF 0.25m. Therefore, ED = V/m. To calculate the magnitude of ED, substitute the known values into the equation: ED = V/m = (1,5V + 0,25m . EF)/0,063m.

To calculate the electron current at location D in the steady state, substitute the known values into the charge conservation equation: i = ED - EF 0.25m = (V/m - 0.25m*EF).

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Complete the following sentence.
A diameter is also a...

Answers

Answer:

A diameter is also a double of radius

A diameter is also a chord

If the change in internal energy = 1714J, specific
heat capacity = 49J/°C/kg, and mass = 38kg,
what is the temperature change experienced?
Give your answer to 2 decimal places.

Answers

Answer:

0.92°C

Explanation:

C = change in Q/m × change in T

so

change in T = change in Q/C ×m

C= 49

m= 38

change in Q= 1714

then

= 1714/49 × 38

= 1714/1862

= 0.92°C

rounded off to 2 d.p

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