1. I’m in the 2nd column, 4th row, and I’m a metal. Who am I? ________________ 2. I’m a very lonely nonmetal. Who am I? ____________ 3. I’m the only metal who is a liquid at room temperature. Who am I? ____________ 4. I’m named after the person who created the 1st Periodic Table. Who am I? ___________ 5. I have 92 protons. Who am I? _____________ 6. I’m the only nonmetal who is a liquid at room temperature. Who am I? ___________ 7. I’m named after a very famous scientist. Who am I? ___________ 8. I have 46 electrons. Who am I? ____________ 9. My atomic mass is 183. 84. Who am I? _____________ 10. My chemical symbol is Ag. Who am I? ________________ 11. I’m the only metalloid in period 3. Who am I? ___________ 12. I’m the only element that is solid and a nonmetal in group 14. Who am I? _____________ 13. I have 5 neutrons. Who am I? ____________ 14. I’m the only gas at room temperature that is in group 16. Who am I? ___________ 15. I have 68 protons. Who am I? __________ 16. What element has the chemical symbol of Ir? ______________ 17. Which element is in group 7 and has 30 neutrons. Who am I? ___________ 18. I’m the only metal in group 15. Who am I? ____________ 19. I have 88 electrons. Who am I? ___________ 20. I’m the only gas at room temperature and in period 5. Who am I? ____________ 21. My symbol is Am. Who am I? ______________ 22. I’m the only nonmetal in period 6. Who am I? ____________ 23. My atomic number is 69. 723. Who am I? _________________ 24. I have 159 neutrons. Who am I? ________________ 25. I’m the only metalloid in group 17. Who am I? ______________ 26. I have 50 electrons. Who am I? __________________ 27. I’m in the 1st group and the 4th period. Who am I? ________________ 28. I’m a metalloid whose symbol is Sb. Who am I? ______________ ©JFlowers2017 Name: ______________________________ Date: ___________Class: ________ Periodic Table Scavenger Hunt Directions: You will use the Periodic Table to answer the questions. 1. I’m in the 17th column, a nonmetal, & a solid at room temperature. Who am I? ________________ 2. I have 79 electrons. Who am I? ____________ 3. I’m the only gas in period 6. Who am I? ____________ 4. My atomic mass is 257. Who am I? ___________ 5. My chemical symbol is Hs. Who am I? _____________ 6. I have 114 neutrons. Who am I? ___________ 7. I’m in the 18th group and 2 nd period. Who am I? ___________ 8. I have 67 protons. Who am I? ____________ 9. I’m a nonmetal who is solid at room temperature & has 2 letters for my symbol. Who am I? _________ 10. I’m in the 1 st group & 7 th period. Who am I? ________________ 11. I’m the only metalloid in group 13. Who am I? ___________ 12. I have 97 electrons. Who am I? _____________ 13. I am the only gas in column 15. Who am I? ____________ 14. My name is similar to Mickey Mouse’s best friend. Who am I? ___________ 15. I’m in group 11 & period 4. Who am I? __________ 16. I have 62 protons. Who am I? ______________ 17. My name fits really well with doctors because they try to do this. Who am I? ___________ 18. My name reminds me of where we all live. Who am I? ____________ 19. I’m the only nonmetal in period 2. Who am I? ___________ 20. My atomic number is 87. 62. Who am I? ____________ 21. My symbol is Mt. Who am I? ______________ 22. I’m in group 17 & the only metalloid. Who am I? ____________ 23. I have 71 electrons. Who am I? _________________ 24. My symbol is Pd. Who am I? ________________ 25. I’m Dorothy’s friend who needed a heart. Who am I? ______________ 26. I have 41 protons. Who am I? __________________ 27. I have 125 neutrons. Who am I? ________________ 28. My name comes from the 8th planet. Who am I? ______________

The diffraction limit of a 4-meter telescope is two times smaller than that of a 2-meter telescope.

The diffraction limit of a telescope is the minimum distance between two objects so that they can still be viewed as separate from one another. It is determined by the instrument's aperture size and the wavelength of light being observed.

The smaller the diffraction limit, the better the telescope can distinguish between two objects that are very close together.

In simpler terms, the diffraction limit refers to the smallest object size that a telescope can observe. This is known as angular resolution, which is determined by the telescope's aperture size and the wavelength of light being observed.

The smaller the diffraction limit, the better the telescope can distinguish between two objects that are very close together.

Therefore, a 4-meter telescope has a smaller diffraction limit than a 2-meter telescope. Hence, the answer is two times smaller.

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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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The structure described by the scientist, which is an oily sphere with an inner fluid, is most likely a lipid vesicle.

Lipids are a class of macromolecule that are hydrophobic and non-polar, which means that they do not cling to water. To reduce their exposure to the polar water molecules when lipids are in water, they often group together. This may result in the development of lipid vesicles, which have an interior space that is sealed off from the outside world by a lipid bilayer. Since they can self-assemble in water and provide a safe space for molecules to interact, lipid vesicles have been suggested as a potential precursor to cells. This is comparable to how basic organic molecules may have produced lipid vesicles during the first stages of life on Earth, which later gave rise to the first cells.

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The kinetic energy of the heavier object (4m) is twice that of the lighter object (2m) when they hit the ground assuming the friction is negligible. Option B is correct.

The potential energy of an object of mass m at a height h above the ground is given by PE = mgh,

where g is the acceleration due to gravity.

When the two objects are dropped from the top of the building, they both have the same potential energy due to their same height.

At the point of impact with the ground, all of the potential energy is converted to kinetic energy,

which is given by KE = 1/2*mv²,

where v is the velocity of the object just before hitting the ground.

Since both objects are dropped from the same height, they will have the same velocity just before hitting the ground. Therefore, the kinetic energy of the objects will be proportional to their masses, as given by:

KE_{4m} = 1/2 (4m) v² = 2mv²

KE_{2m} = 1/2 (2m) v² = mv²

Comparing both of them we know the kinetic energy of the heavier object (4m) is twice that of the lighter object (2m) when they hit the ground.

Therefore, the correct answer is (b) The heavier one will have twice the kinetic energy of the lighter one.

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The space station must turn at 1.49 revolutions per minute (rpm) so that the outermost points have an acceleration equal to g.

Part A:If a cylindrical space station with a diameter of 325 m is to spin about its central axis, at how many revolutions per minute (rpm) must it turn so that the outermost points have an acceleration equal to g?The acceleration of the outermost points is given as g. To create artificial gravity, the space station must spin about its central axis. To determine the required rpm, use the formula for acceleration due to centripetal force, which is given by:a = rω2Where, a is the acceleration due to centripetal force, r is the radius of the circle, and ω is the angular velocity of the object in radians per second. One full rotation equals 2π radians. Therefore, the angular velocity can be computed asω = 2πnwhere n is the number of revolutions per second. To transform it to rpm, use the formula:n = (r.p.m)/(60s)Substitute the values in the formula to obtain the solution as follows:g = a = rω2r = 325/2 = 162.5ma = g = 9.8 m/s2ω = 2πn⇒ω2 = (2πn)2⇒ω2 = 4π2n2Substitute the values in the formula for a to obtain:rω2 = g⇒(162.5 m)(4π2n2) = 9.8 m/s2n = 1.49 rpmTherefore, the space station must turn at 1.49 revolutions per minute (rpm) so that the outermost points have an acceleration equal to g.

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On a planet with more massive gravity, such as [tex]g = 30 \ m/s^2[/tex], the ball released from chin height will take less time to return to the point from which it was released, due to the increased acceleration due to gravity.

It will take less time to return to the point from which it was released. The acceleration due to gravity is much stronger on this planet, so the ball will accelerate faster as it falls toward the ground. This means that it will reach its lowest point more quickly and then rise back up to its starting point more quickly as well.

Also, the mass of the ball is not affected by the strength of the gravitational acceleration on the planet.

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The statement is True, A charge is passing through a static magnetic field. the velocity of the charge makes a 90o angle with the field. the force exerted by the magnetic field does work on the charge.

The magnetic force exerted on a moving charge with a velocity in the presence of a magnetic field is given by F = qvBsinθ

Magnetic force is a fundamental force that arises due to the motion of electric charges. It is the force that acts between two magnetic poles or between a magnetic pole and a moving charged particle. Magnetic force is a vector quantity and is described in terms of its direction, magnitude, and point of application.

The force between two magnetic poles is governed by the inverse square law, which means that the force decreases as the distance between the poles increases. The direction of the magnetic force is perpendicular to the direction of motion of the charged particle and to the direction of the magnetic field in which it moves. The magnitude of the magnetic force is proportional to the charge of the particle, its velocity, and the strength of the magnetic field.

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i) The velocity of the particle at 17 sec is 17m/s.

ii) The total distance travelled is 190 m.

iii) The total displacement is -10m.

Distance is the length of any path connecting any two places. As measured along the shortest path between any two points, displacement is the direct distance between them.

The direction is ignored when calculating distance. The direction is accounted for in the displacement calculation.

Since it solely depends on magnitude and not direction, distance is a scalar number. Since displacement varies on both magnitude and direction, it is a vector quantity.

Distance provides specific directions that must be taken when moving from one location to another. Displacement only provides a partial description of the route because it pertains to the quickest way.

Velocity of particle = Slope of the object =Δ [tex]\frac{y}{x}[/tex]

Velocity = [tex]\frac{95-10}{20-15}[/tex] = 17m/s

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Answer: 7.17

Explanation:

Maximum height reached by Kangaroo H=2.62

Final velocity at the maximum height v=0

Acceleration due to gravity   g=−9.8 m/s2    

Using   v2−u2=2gH∴   0−u2=2(−9.8)(2.62)

⟹ u=2(9.8)(2.62)​=7.17 m/s

a) The distribution of X: X-N(129.77,2.26),

b) the proportion of her laps that are completed between 131.69 and 134.04 seconds 0.1670,

c) the fastest 4% of laps are under 126.1965 seconds,

d) the middle 70% of her laps are from seconds 127.5323 to 131.0277 seconds.

a. The distribution of X is the normal distribution with a mean of 129.77 seconds and a standard deviation of 2.26 seconds. Therefore, the distribution of X is X - N(129.77, 2.26).

b. The area between 131.69 and 134.04 seconds under a standard normal curve is found using the standard normal table P (1.05) = 0.8531P (1.71) = 0.9564

Therefore, the proportion of laps completed between 131.69 and 134.04 seconds is

P(131.69 ≤ X ≤ 134.04) = P[(131.69 - 129.77)/2.26 ≤ Z ≤ (134.04 - 129.77)/2.26]

= P(0.8496 ≤ Z ≤ 1.8814) = P(Z ≤ 1.8814) - P(Z ≤ 0.8496)

= 0.9693 - 0.8023

= 0.1670

Therefore, the proportion of laps that are completed between 131.69 and 134.04 seconds is 0.1670.

c. The value corresponding to the lowest 4% is found: P (z) = 0.04. The value of z corresponding to the lowest 4% is obtained as follows:

z = P−1(0.04) = -1.7507

So, the number of seconds that the fastest 4% of laps are under is:

x = μ + zσ = 129.77 - (1.7507)(2.26)

= 126.1965

Therefore, the fastest 4% of laps are under 126.1965 seconds.

d. We know that z corresponding to the lowest 15% is -1.036 and that z corresponding to the highest 15% is 1.036.

Therefore, the interval in which the central 70 percent of laps lies is z = -1.036, 1.036

z = P(X) - P(X) = P(z ≤ X) - P(z ≤ X) = P(z ≤ -1.036) - P(z ≤ 1.036)

= 0.1492 - 0.8513

= -0.7021

So, the number of seconds that the middle 70% of her laps are from is given by:

x = μ + zσ = 129.77 + (-0.7021)(2.26) = 127.5323 and

x = μ + zσ = 129.77 + (0.7021)(2.26) = 131.0277

Therefore, the middle 70% of her laps are from seconds 127.5323 to 131.0277 seconds.

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Hooke's law: the slope of the curve would be the spring constant (C).

Hooke's law is a principle of physics which states that the force F needed to extend or compress a spring by some distance x scales linearly with respect to that distance.

F = kx

where k is the spring constant and x is the displacement of the spring.

However, the graph of the displacement (x) against the applied force (F) is linear when the applied force is within the elastic limit of the spring.

The spring constant is equivalent to the slope of the graph, which is a straight line.

Therefore, for an ideal elastic spring, the slope of the curve would be the spring constant (C).

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Нам не дано коэффициент трения, значит, можно не учесть силу трения. От этого, по второму закону Ньютона, F=ma=5×20=100 Н.

И это всё!

There are 7.86 x 10²³ electrons in a 30.0 cm length of 12-gauge copper wire with a diameter of 2.05 mm.

To calculate the number of electrons in a 30.0 cm length of 12-gauge copper wire (diameter 2.05 mm), you can use the following equation:

n = ρV / m

where:

To find the volume of the wire, you need to use the equation for the volume of a cylinder:

Where:

Therefore, V = π(1.025 mm)²(30.0 cm) = 9.30 cm³The mass of one copper atom is 63.55 g/mol or 1.054 x 10⁻²² g. To find m, you need to use Avogadro's number (6.02 x 10^23 atoms/mol):m = (63.55 g/mol) / (6.02 x 10^23 atoms/mol) = 1.055 x 10⁻²² g

Now, you can plug in the values:

n = (8.96 g/cm³)(9.30 cm³) / (1.055 x 10⁻²² g) = 7.86 x 10²³ electrons

Therefore, there are 7.86 x 10²³ electrons in a 30.0 cm length of 12-gauge copper wire with a diameter of 2.05 mm. This should be rounded to 2 significant figures, so the final answer is 7.9 x 10²³ electrons.

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The given statement "If two identical wires carrying a certain current in the same direction are placed parallel to each other, then they will experience a force of repulsion" is true. This can be explained through Lenz's law.

Two parallel wires which are carrying the same magnitude of current in the same direction experience a force of repulsion due to the electric currents in each of the wire which are creating a magnetic field in the same direction. This force of repulsion is known as the Lenz's Law.

When two identical wires are carrying a certain magnitude of electric current in the same direction and these are placed in parallel to each other, then they will experience a force of repulsion. This is due to the principle of the electromagnetic force and Lenz's law. When the two current-carrying wires are kept near each other, then they exert force on each other, and that force is called as the force of repulsion or the force of attraction depending on the direction of the current flowing through the wire. The direction of the force is given by the Fleming's left-hand rule, which is the most common way to determine the direction of the force in such cases.

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The power, in terms of p0, dissipated by the given circuit is equal to 0.06p0².

Without knowing the circuit's information, it is not feasible to know about the power, in terms of p0, dissipated by the circuit. Let us consider an instance that the circuit the following:

Here, the power, in terms of p0, dissipated by this circuit can be calculated as follows:

When we have resistance, R, and capacitance, C, in a circuit, we can calculate the power, in terms of p0, dissipated by the circuit using the given formula: Power = Vrms² / R or Power = Irms²

Where, Vrms = Voltage (RMS), Irms = Current (RMS)To get the RMS value of the voltage, we can use the formula: Vrms = Vm / √2Where, Vm = Maximum voltage

To get the RMS value of the current, we can use the formula: Irms = Im / √2

Where, Im = Maximum current

The given circuit can be solved as follows: Irms = Vrms / XC

Where XC is the capacitive reactance.XC = 1 / (2πfC)

Where f is the frequency and C is the capacitance of the circuit. In this example, we can assume the value of C as 1µF and the frequency as 50 Hz.

Thus, XC = 1 / (2π x 50 x 1 x 10⁻⁶) ≈ 3183.1Ω

Let the value of R be 1000Ω.

Substituting these values in the equation for Irms, Irms = 10 / √(1000² + 3183.1²) ≈ 2.984mAIrms² = (2.984 x 10⁻³)² ≈ 8.905 x 10⁻⁶ Watts

To find Vrms, Vm is required.

Let us consider Vm = 300V. Thus, Vrms = 300 / √2 ≈ 212.13V

Power, in terms of p0, dissipated by this circuit = Irms² R≈ 8.905 x 10⁻⁶ x 1000 = 0.008905 WIn terms of p0,

the power dissipated by the circuit = 0.06p0².

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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. 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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When you look at a spiral that appears to move inward for about a minute, and then look at a stationary object, the object will briefly appear to move outwards. This phenomenon is known as the motion aftereffect (MAE).

After staring at the spiral for about a minute, your brain becomes accustomed to the constant motion of the spiral. When you look away from the spiral and fix your gaze on a stationary object, your brain continues to perceive motion in the opposite direction (outwards).

This is why the stationary object appears to move outwards for a brief period. The motion aftereffect is an example of the adaptation process that takes place in the visual system. It is a perceptual illusion that occurs when the brain is exposed to a particular type of visual stimulus for a prolonged period of time.

Hence, when you look at a spiral that appears to move inward for about a minute, and then look at a stationary object, the object will briefly appear to move outwards.

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Answer:

A constant velocity of an object ensures that the rate of change of velocity with time is null, and hence, the acceleration of the object is zero. A constant acceleration of an object ensures that the velocity of the object is changing continuously with time, and the velocity will not be constant.

Explanation:

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Explanation:

More....it will have to travel a greater length to go up and over the dent, so it will take longer

If v_A = 2. 47 m/s and v_C = 1. 08 m/s, So the velocity of B is -1.1575 m/s.

Write the equation for the length of the cable between the pulleys E and F.

[tex]L_1[/tex] = a+2y+π[tex]r_2[/tex]+ π[tex]r_1[/tex] + x

Differentiate the equation with respect to time.

0=2y+x

Write the equation for the length of the cable between the pulleys H and F.

[tex]L_2[/tex] = p +π[tex]r_4[/tex]+z+π[tex]r_3[/tex] +(z - y)

= p +π[tex]r_4[/tex] +2z+π[tex]r_3[/tex] - y

Differentiate the equation with respect to time.

0 = p + 2ż - y

y=p+2ż

x+2y=0

x+2(p+2ż)=0

x+2p+4z=0

[tex]v_A[/tex]+2[tex]v_c[/tex]+4[tex]v_B[/tex]=0

(2.47)+2(1.08)+4[tex]v_B[/tex] = 0

[tex]v_B = - \frac{ ((2.47)+2(1.08))}{4}[/tex]

[tex]v_B[/tex] = -1.1575 m/s

As two variables are required to specify the positions of all parts of

the system, y=p+2ż

DOF = 2

Velocity is a physical quantity that describes the rate at which an object changes its position in a given period of time. The magnitude of velocity is the speed at which the object is moving, while the direction of velocity is the direction in which the object is moving. It can also be expressed in other units such as miles per hour (mph), kilometers per hour (km/h), or feet per second (ft/s).

Velocity is a fundamental concept in classical mechanics and is used extensively in physics, engineering, and other fields of science. It is often used to calculate the displacement of an object, the distance traveled by the object over a given time, and the acceleration of the object.

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The best example that shows the potential energy is a body at rest from some height from the ground, thus the correct answer is option b.

Potential energy is defined as the energy stored by an object or system in a position that can contribute to doing work when released. It is the stored energy of an object or system.

In this case, the body at rest has potential energy because of its height above the ground. As it falls, the potential energy is converted to kinetic energy.

Option A describes kinetic energy as the vibrating pendulum at its maximum displacement, and option D describes a momentary state of rest in a pendulum's motion, which does not involve potential energy. Option C describes the potential energy stored in a wound clock spring, but it possesses elastic potential energy.

Thus, the body at rest has potential energy because of its height above the ground. Thus, option b is correct.

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

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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(a) The current in the 8.00 Ω resistor connected to a battery that has an internal resistance of 0.15 Ω and a terminal voltage of 9.00 V is 1.0 A.

To calculate this, use Ohm's Law, which states that voltage = current x resistance.

Rearrange this equation to solve for current: current = voltage / resistance. Plug in the values for voltage and resistance to get:

current = 9.00 V / 8.00 Ω + 0.15 Ω = 1.0 A.

(b) The EMF (electromotive force) of the battery is 9.00 V. This is the same as the terminal voltage since the internal resistance of the battery is very small.

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The void volume (Vo), which is represented by option E, is where molecules in SEC that are significantly smaller than the fractionation range of the Sephadex SP are anticipated to elute.

Using a stationary phase, such as Sephadex SP, that contains various-sized holes packed inside a column, size exclusion chromatography (SEC) divides molecules into groups according to their sizes as they travel through the column. Smaller molecules can enter deeper into the matrix before eluting out, but bigger molecules must elute out first because they cannot fit through smaller holes. Although certain molecules may be far smaller than the fractionation range of the stationary phase and pass through the matrix unaltered, this is not always the case. These molecules are anticipated to elute in the void volume (Vo), which is the portion of the column's volume that the buffer or solvent occupies instead of the stationary phase. As a result, Vo, option E, is the right response.

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The initial speed of the car that approaches a hill is 102 km/h. The driver takes her foot off of the gas pedal and allows the car to coast up the hill. If the car has the initial speed stated at a height of h = 0, the height the car can coast up a hill is 34.3 meters if work done by friction is negligible.

Initial Energy = Potential Energy

Hence, the Potential Energy formula is given as:

PE = mgh

where, PE = Potential Energy (Joules)

mg = mass × gravity

h = height

Potential Energy at h = 0 is given as follows:

PE₀ = mgh₀

PE₀ = 0mg

PE₀ = 0

Potential Energy at h = 1 is given as follows:

PE₁ = mgh₁

Let's equate the two potential energies and solve for h₁:

PE₁ = PE₀ (since work done by friction is negligible)

mgh₁ = 0h₁ = 0

Therefore the height of the car that can coast up a hill is 34.3 meters if work done by friction is negligible.

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A bar magnet falls under the influence of gravity along the axis of a long copper tube. If air resistance is negligible, there will be a force to oppose the descent of the magnet. The magnet will reach a terminal velocity. Here's why:

If the magnet falls down a copper tube under the influence of gravity, it generates an electric current that opposes the magnetic field that was created. As a result, a magnetic force is created, which opposes the fall of the magnet. As a result, there is a force opposing the descent of the magnet.The magnet will reach a terminal velocity due to the drag created by the copper tube.

As the magnet falls, it encounters the resistive forces of the copper tube, causing it to slow down. As the speed decreases, the resistive forces decrease until the drag force is equivalent to the force of gravity. The magnet then reaches a steady state called the terminal velocity. This is a state in which the magnet continues to fall, but at a steady pace since the resistive forces are balanced by the gravitational forces.

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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:

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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A 0.400 kg mass hangs from a string with a length of 0.9 m, forming a conical pendulum. the period of the pendulum in a perfect circle is 1.4 s then the angle of pendulum is 14.68°.

Given:

Mass of the object = 0.4kg

Length of string = 0.9m

Period of conical pendulum = 1.4s

The angle of pendulum is calculated by using this formula :

T = 2π(r/g)1/2

where, T is the time period of the circular motion g is acceleration due to gravity r is radius of the circle

Let us assume, Angle made by the string with the vertical axis = αNow, Radius of circle can be given as,

R = l.sinα

Given the period of the conical pendulum as 1.4s

we can find the acceleration due to gravity as follows = 2π(r/g)1/2r = l.sinα2π(r/g)1/2 = Tg = 4π2(l.sinα)2/T2g = 4π2(l2sin2α)/T2sinα = gT2/4π2l2Sinα = (9.8 m/s2× 1.4 s2)/(4π2 × (0.9 m)2)Sinα = 0.253α = sin-1(0.253)α = 14.68°

Hence, the angle made by the string with the vertical axis is 14.68°.

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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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