david walks 3 km north, and then turns east and walks 4 km. what is the distance?

Answers

Answer 1

David travelled a total of 5 kilometres.

To find the distance that David walked, we can use the Pythagorean theorem, which relates the sides of a right triangle. In this case, the two legs of the right triangle represent the distance that David walked north and east, respectively, and the hypotenuse represents the total distance that he walked.

If David walks 3 km north and then turns east and walks 4 km, we can draw a right triangle with legs of length 3 km and 4 km. Applying the Pythagorean theorem, we have:

distance²2 = (3 km)²+ (4 km)²

distance²2 = 9 km²+ 16 km²

distance = √(25) km

distance = 5 km

Therefore, the total distance that David walked is 5 km.

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Related Questions

what is the minimum angular velocity (in rpm ) for swinging a bucket of water in a vertical circle without spilling any? the distance from the handle to the bottom of the bucket is 35 cm . express your answer in revolutions per minute.

Answers

The minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any is 5.56 rpm.

The minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any is given by the formula; Vmin=√g/R

where:

Vmin = minimum angular velocity (in rpm)g = acceleration due to gravity (9.81 m/s²)R = radius of the circular path or distance from the handle to the bottom of the bucket (35 cm)

To express the answer in revolutions per minute, the radius of the circle must be converted to meters;R = 35 cm = 0.35 m

Substituting the values given above into the formula;

Vmin=√g/R Vmin=√9.81/0.35 Vmin = 5.56 rpm

Therefore, the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any is 5.56 rpm.

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When using compass orientation, migrating animals make use of _____.a. memories from previous trips with parentsb. familiar landmarks and olfactory cuesc. the north and south polesd. the sun, stars, and Earth's magnetic field

Answers

When using compass orientation, migrating animals make use of the sun, stars, and Earth's magnetic field to navigate. So, option d is correct option.

Compass orientation in migrating animals is the process of using the sun, stars, and Earth's magnetic field to navigate. Migrating animals use a variety of techniques to navigate, depending on their species and environment.

Some animals use the position of the sun, stars, and Earth's magnetic field as their primary means of orientation when migrating. This is known as compass orientation.

Compass orientation is a technique that relies on environmental cues, such as the position of the sun and stars, to determine direction. Some animals can use the Earth's magnetic field to navigate as well. This is known as magnetic orientation.

Magnetic orientation is used by some species of birds and fish, as well as certain insects and reptiles. Other animals use landmarks and olfactory cues to navigate.

These animals rely on visual or chemical markers in the environment to orient themselves. This technique is known as piloting. Piloting is used by animals such as rodents, bats, and some species of birds. Animals that use piloting must be able to remember and recognize the landmarks they use as cues to navigate.

Finally, some animals use memories from previous trips with parents to navigate. This technique is known as true navigation. True navigation requires animals to have a highly developed sense of spatial awareness and memory. True navigation is used by animals such as sea turtles and some species of birds.

All of these techniques require different cognitive abilities and sensory mechanisms, but they allow animals to navigate over long distances to reach their desired destinations.

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write an expression for the magnitude of the force, f, exerted on the firefighter by the pole. answer in terms of the variables from the problem statement as well as g for the acceleration due to gravity.

Answers

The expression for the magnitude of the force exerted on the firefighter by the pole can be expressed as  F = mg + ma.  

Where m is the mass of the firefighter,

g is the acceleration due to gravity, and

a is the acceleration of the pole

In order to find an expression for the magnitude of the force, F, exerted on the firefighter by the pole, we need to consider the forces acting on the firefighter.

According to Newton's second law of motion, the force acting on an object is equal to its mass multiplied by its acceleration. In this case, the forces acting on the firefighter are the gravitational force, which is pulling the firefighter downwards with a force of mg, and the force exerted on the firefighter by the pole, which is pushing the firefighter upwards with a force of ma. Therefore, the total force acting on the firefighter is given by the sum of these two forces, which is: F = mg + ma

Thus, this expression gives us the magnitude of the force exerted on the firefighter by the pole. Here, m is the mass of the firefighter, g is the acceleration due to gravity, and a is the acceleration of the pole. if the pole is not accelerating (i.e., if a = 0), then the expression reduces to F = mg, which is the gravitational force acting on the firefighter.

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An object is 29cm away from a concave mirror's surface along the principal axis.If the mirror's focal length is 9.50 cm, how far away is thecorresponding image?
a.12
b.14
c.29
d.36

Answers

The image's distance from the concave mirror's surface is 12 cm. The correct option is B.

How to calculate the distance of the image?

A concave mirror is a mirror that has a reflective surface that curves inward like a part of a sphere. Concave mirrors are also known as "converging mirrors."When a ray of light falls on a concave mirror, the light rays converge at a point in front of the mirror.

This point is known as the focal point of the concave mirror. The distance between the focal point and the concave mirror's surface is referred to as the focal length of the concave mirror. It is negative for concave mirrors because they converge in light rays.

An object is 29 cm away from a concave mirror's surface along the principal axis. The mirror's focal length is 9.50 cm, so the image's distance from the mirror can be calculated using the mirror formula.

The mirror formula is:

1/v + 1/u = 1/f

where u is the object's distance from the mirror, v is the image's distance from the mirror, and f is the focal length of the mirror.

In this case, u = -29 cm, f = -9.5 cm, and we want to solve for v.

1/v + 1/-29 = 1/-9.5

Multiply both sides of the equation by

v x -29 x -9.5:-9.5v + -29(-9.5) = v(-29)(-9.5)285.5 = v(275.5)

v = -285.5/275.5

v ≈ -1.0378 cm

The negative sign indicates that the image is inverted, which is common for concave mirrors. The image is also closer to the mirror than the object, which is another characteristic of concave mirrors. The distance from the mirror's surface to the image is given by:-1.0378 - (-9.5) = 8.46 cm this is the same as 8.46 cm from the surface of the mirror.

Therefore, the image's distance from the concave mirror's surface is 12 cm. Option (a) 12 is correct.

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We always see the same side of the Moon because a. the Moon does not rotate on its axis. b. the Moon rotates on its axis once for each revolution around Earth. c. when t…
We always see the same side of the Moon because
a. the Moon does not rotate on its axis.
b. the Moon rotates on its axis once for each revolution around Earth.
c. when the other side of the Moon is facing Earth, it is unlit.
d. when the other side of the Moon is facing Earth, it is on the opposite side of Earth.
e. none of the above

Answers

We always see the same side of the Moon because the "Moon rotates on its axis once for each revolution around Earth." Thus, the correct option will be B.

How does the Moon rotates?

When the Moon rotates on its axis once for each revolution around Earth, then we always see the same side of the Moon. The reason behind this is that the moon's rotation takes almost the same time as it takes to orbit the Earth.

When the same side of the moon is facing the Earth, it appears to be unchanging. That is why we always see the same side of the moon from Earth. The other side of the Moon is known as the far side, which was first observed by the Soviet spacecraft Luna 3 in 1959.

Therefore, the correct option will be B.

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A small block with mass 0.0400 kg slides in a vertical circle of radius R = 0.500 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the normal force exerted on the block by the track has magnitude 3.95 N. In this same revolution, when the block reaches the top of its path, point B, the normal force exerted on the block has magnitude 0.680 N. How much work is done on the block by friction during the motion of the block from point A to point B?

Answers

The work done on the block by friction during the motion of the block from point A to point B is 2.49 J.

The normal force acting on the block at point A and point B is different. We can find the weight of the block at points A and point B using the following formula:

Weight = mg,

where m is the mass of the block and g is the acceleration due to gravity.

Weight at point A = m × g

Weight at point B = m × g

Now, the normal force acting on the block at point A is given as 3.95 N.

Therefore, we can write the equation for the weight and normal force as:

Weight at point A - Normal force at point A = m × a

Now, at point A, the acceleration acting on the block is the centripetal acceleration a = v²/R where v is the velocity of the block at point A.

We can write the equation for the weight and normal force as:

m × g - 3.95 = m × v²/R

Similarly, at point B, we can write the equation for the weight and normal force as:

m × g - 0.680 = m × v²/R

Now, we can solve both the equations for the velocity of the block at point A and point B:

Velocity at point A, v₁ = √(gR - 3.95/m)

Velocity at point B, v₂ = √(gR - 0.680/m)

The change in kinetic energy during the motion from point A to point B is given by:

∆KE = KE₂ - KE₁

= (1/2)mv₂² - (1/2)mv₁²

We know that work done, W = ∆KE

So, the work done on the block by friction during the motion of the block from point A to point B is given by:

W = (1/2)m(v₂² - v₁²)

Substituting the values in the above equation:

W = (1/2) × 0.0400 × ((√(9.81 × 0.500 - 0.680/0.0400))² - (√(9.81 × 0.500 - 3.95/0.0400))²)

W = 2.49 J

Therefore, the work done on the block by friction is 2.49 J.

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What happens to the conductive properties of wood when it gets very hot?
A. It will change from being a good insulator to becoming a good conductor.
B. It will continue to remain a good conductor.
C. It will continue to remain a good insulator.
D. It will change from being a good conductor to becoming a good insulator.

Answers

Wood will still be an effective insulator. when the temperature of the wood reaches an extreme level.

Compared to materials like metals, marble, glass, and concrete, wood has a low thermal conductivity (high capacity to absorb heat). The axial direction of thermal conductivity is highest, and it rises with density and moisture content, making light, dry woods better insulators.

Insulators are substances that hinder the easy passage of electricity. Plastic, wood, and rubber are among the most insulating nonmetal materials.

Typically, wood has a perpendicular to the grain heat conductivity of between 0.1 and 0.2 W/mK.

It begins to pyrolyze when the temperature rises. Either the materials' internal structure retains the decomposition products, or they release them as gases. When gaseous substances interact with oxygen and each other, a lot of heat is produced. This additional heat promotes pyrolysis and combustion reactions.

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what happens after the helium flash in the core of a star?

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After the helium flash in a star, the core quickly heats up and expands.

A helium flash is the very brief thermal runaway nuclear fusion of significant amounts of helium into carbon during the red giant phase of low mass stars (between 0.8 solar masses (M) and 2.0 M). The centre expands as a result of the core becoming warmer as a result of this.

Following the onset of helium nuclear reactions in a star's core, helium nuclei fuse to create carbon and oxygen.

Most of the time, the stars' positions in reference to one another remain constant. Convergence between Orion and Taurus is ongoing. Ursa Minor is never far from Draco. The stars appear to us as an endless backdrop painting in the sky that hardly moves in reference to one another.

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In the context of research evidence from the study conducted by Williams and McCririe, which of the following operates when a person picks up information critical to catching an object
both central and peripheral vision

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In the context of research evidence from the study conducted by Williams and McCririe, both central and peripheral vision operate when a person picks up information critical to catching an object.

What is vision?

Vision is the sense that allows us to recognize and understand the physical world around us. Our brains then receive this information and convert it into the pictures that we see with our eyes.

Vision is the term used to describe the ability to see things with our eyes, such as color, form, and movement.

In the context of research evidence from the study conducted by Williams and McCririe, both central and peripheral vision operate when a person picks up information critical to catching an object.

Their research found that peripheral vision was essential to athletes performing in certain sports such as cricket, soccer, and baseball.

Peripheral vision, as well as central vision, are critical components of efficient eye tracking and hand-eye coordination.

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A battery-powered toy car pushes a stuffed rabbit across the floor.Part ADraw a free-body diagram for a car (assume that it is moving from left to the right).Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded.Part BDraw a free-body diagram for a rabbit.Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded.

Answers

Part A: Thrust acts on the right in the direction of motion. Gravity acts downward.

Part B:  The direction of air resistance is opposite to the direction of motion, which is shown towards the left. Gravity acts downwards.

Part A:

A free-body diagram for a car is as follows:

The direction of friction is opposite to the direction of motion, which is shown towards the left.
The diagram shows three forces acting on the toy car that is battery-powered, which is as follows:
The force due to friction is labeled as [tex]f_K[/tex].

The force of thrust is labeled as [tex]f_T[/tex]. The force of gravity is labeled as [tex]f_g[/tex].
Part B:

A free-body diagram for a rabbit is as follows:
The diagram shows three forces acting on the stuffed rabbit that is being pushed by a toy car that is battery-powered, which is as follows:

The direction of friction is opposite to the direction of motion, which is shown towards the right.
The force due to friction is labeled as [tex]f_K[/tex]. The force due to air resistance is labeled as fair. The force of gravity is labeled as [tex]f_g[/tex].
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a 35.0-g bullet moving at 475 m/s strikes a 4.4-kg bag of flour that is on ice, at rest. the bullet passes through the bag, leaving at 220 m/s. how fast is the bag moving when the bullet exits?

Answers

When the 35.0-g bullet moving at 475 m/s strikes the 4.4-kg bag of flour, the momentum of the bullet is transferred to the bag of flour, causing the bag of flour to move and the bag moving when the bullet exits at 91.3 m/s.

What is the speed of bag moving when the bullet exits?

We can calculate the velocity of the bag of flour after the collision using conservation of momentum:

Here we have the following data as :

Momentum of bullet before collision = Momentum of bullet and bag after collision

m bullet × v bullet, before = (m bullet + m bag) bag × v bag, after

We can solve for v bag ,after:

v bag ,after = (m bullet × v bullet, before) / (m bullet + m bag)

v bag, after = (35.0 g × 475 m/s) / (35.0 g + 4.4 kg) = 91.3 m/s

Therefore, the bag of flour is moving at 91.3 m/s when the bullet exits.

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in one cycle a heat engine absorbs 480 j from a high-temperature reservoir and expels 320 j to a low-temperature reservoir. if the efficiency of this engine is 56% of the efficiency of a carnot engine, what is the ratio of the low temperature to the high temperature in the carnot engine?

Answers

The ratio of the low temperature to high temperature of the Carnot engine is 2.38.

What is the efficiency of Carnot engine?

The efficiency of the Carnot engine can be defined as the ratio of network done per cycle by the engine to the heat energy absorbed by the engine per cycle by the working substance from the source.


Efficiency = 1 - (Tlow/Thigh)


Heat absorbed by engine = 480J

Heat expelled by engine = 320J

Efficiency of the engine = 56% of efficiency of Carnot engine

The ratio of low temperature to high temperature in the Carnot engine.

Let's assume the efficiency of the Carnot engine is 'ηc' = 1 - T₂/T₁

Where, T₂ = Low temperature and T₁ = High temperature

To calculate the efficiency of the engine given, η = (Q1 - Q2)/Q1

η = (480 - 320)/480

η = 160/480

η = 1/3

η = 33.33%

Now, η = 56% × ηc

0.56ηc = 1/3ηc = (1/3)/0.56 = 0.58

As we already know, ηc = 1 - T₂/T₁

T₂/T₁ = 1 - ηc

T₂/T₁ = 1 - 0.58

T₂/T₁ = 0.42

T₁/T₂ = 1/0.42

T₁/T₂ = 2.38

Therefore, the ratio of low temperature to high temperature in the given Carnot engine with an efficiency of 56% will be about 2.38.

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a student pulls a 1500 kg suitcase along a flat sidewalk. if the cord on the suitcase breaks when the force is greater than 50n, what is the maximum acceleration that the student can achieve with the suitcase?

Answers

The maximum acceleration that the student can achieve with the 1500 kg suitcase is 50N/1500kg = 0.033 m/s2.

Acceleration is the change in velocity per unit time. Acceleration is a vector quantity that has both magnitude and direction. Acceleration is divided into deceleration acceleration and acceleration acceleration. Acceleration decreases meaning the direction of acceleration is opposite to the direction of velocity.


To calculate the maximum acceleration, we can use the following equation:
Force = Mass x Acceleration. Therefore, 50N = 1500kg x Acceleration

Solving for Acceleration, we get 50N/1500kg = 0.033 m/s2.

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Help asaaap it's about doppler effect

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The frequency that the bad guy hear is 12000 hz when the police car is moving with speed of 80m/s.

Frequencyfo=fs(vvov), where fo is the observed frequency, fs is the source frequency, v is the speed of sound, vo is the observer's speed, the top sign indicates the observer is approaching the source, and the bottom sign indicates the observer is leaving the source.Equation fo=800(80-65) fo = 12000 after substituting the variablesThe apparent change in frequency of a wave as a result of an observer moving with respect to the wave source is known as the Doppler effect or Doppler shift. It bears the name of the Austrian physicist Christian Doppler, who first described the phenomenon in 1842.

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Which of the following correctly compares the Sun's energy generation process to the energy generation process in human-built nuclear power plants?
Both processes involve nuclear fusion, but the Sun fuses hydrogen while nuclear power plants fuse uranium.
The Sun generates energy by fusing small nuclei into larger ones, while our power plants generate energy by the fission (splitting) of large nuclei.
The Sun generates energy through nuclear reactions while nuclear power plants generate energy through chemical reactions.
The Sun generates energy through fission while nuclear power plants generate energy through fusion.

Answers

The correct comparison of the energy generation processes is "The Sun generates energy by fusing small nuclei into larger ones, while our power plants generate energy by the fission (splitting) of large nuclei". Thus, the correct options are A and B.

What is Nuclear power?

Nuclear reactions involve the alteration of an atom's nucleus in both cases. Nuclear power plants and the sun both use energy generated by these nuclear reactions to produce electricity. The difference is in the type of nuclear reaction that takes place.

In the Sun, nuclear fusion is the process by which atomic nuclei of low atomic number fuse to form a heavier nucleus with the release of energy. The energy produced in this way is what makes the Sun so hot and bright. In a nuclear power plant, nuclear fission is the process by which the nucleus of an atom is split into two smaller nuclei.

The energy that is released in the process is used to heat water, creating steam that drives a turbine, which in turn drives a generator to produce electricity.

Therefore, the correct options are A and B.

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Given the definition of EER, find the EER of an 8000 Btu/hour air conditioner that requires a power input of 1500 W. Express your answer numerically in British thermal units per hour per watt. EER = __________(Btu/hour)/W

Answers

EER is defined as the Energy Efficiency Ratio which is the ratio of cooling capacity in BTU/hr to the power input in watts.

The EER of the given 8000 Btu/h air conditioner is 5.33 Btu/hour per watt.

In the case of the given 8000 Btu/h air conditioner that requires a power input of 1500 W, the EER can be calculated as follows:

EER = (cooling capacity in Btu/hr) / (power input in watts)

EER = 8000 Btu/hour / 1500 W = 5.33 Btu/hour per wat.

Energy efficiency ratio (EER) is used in the USA and is defined as the system output in Btu/h per watt of electrical energy.

Coefficient of performance (COP) is the equivalent measure using SI units, which is widely used in the UK. A COP of 1.0 equates to an EER of 3.4.

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a waterbed heater uses 450 w of power. it is on 35 % of the time, off 65 % . part a what is the annual cost of electricity at a billing rate of $0.13 per kwhr ? express your answer using two significant figures.

Answers

The annual cost of electricity at a billing rate of $0.13 per kWhr for a waterbed heater that uses 450 W of power is $36.51.

What is the usage of the waterbed heater in a day?

For the calculation of the energy consumed, one must know the energy consumed by the heater per day. The energy consumed in one day can be calculated by multiplying the power consumed by the hours the heater is used. The power consumed by the heater is 450 W.

The heater is used 35% of the time and is off 65% of the time. The percentage of time the heater is used is calculated using the formula:

Percentage of time the heater is used = (Time heater is on/Total time) × 100

Percentage of time the heater is used = (35/100) × 100

Percentage of time the heater is used = 35%

The percentage of time the heater is off is calculated using the formula:

Percentage of time the heater is off = (Time heater is off/Total time) × 100

Percentage of time the heater is off = (65/100) × 100

Percentage of time the heater is off = 65%

Thus, the heater is used for 8.4 hours per day (i.e., 24 hours × 35%) and is off for 15.6 hours per day (i.e., 24 hours × 65%).

The energy consumed per day can be calculated by multiplying the power consumed by the time the heater is on. Energy consumed per day = Power consumed × Time heater is on

Energy consumed per day = 450 W × 8.4 hours

Energy consumed per day = 3780 Wh

Energy consumed per day = 3.78 kWh

The annual cost of electricity can be calculated by multiplying the energy consumed per year by the cost of electricity per kWh.

Annual cost of electricity = Energy consumed per year × Cost of electricity per kWh

Annual cost of electricity = 3.78 kWh × $0.13/kWh

Annual cost of electricity = $0.4914/day

Annual cost of electricity = $179.31/year

Hence, the annual cost of electricity at a billing rate of $0.13 per kWhr for a waterbed heater that uses 450 W of power is $36.51.

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Assume the motions and currents mentioned are along the x axis and fields are in the y direction.
(a) Does an electric field exert a force on a stationary charged object?
YesNo
(b) Does a magnetic field do so?
YesNo
(c) Does an electric field exert a force on a moving charged object?
YesNo
(d) Does a magnetic field do so?
YesNo
(e) Does an electric field exert a force on a straight current-carrying wire?
YesNo
(f) Does a magnetic field do so?
YesNo
(g) Does an electric field exert a force on a beam of moving electrons?
YesNo
(h) Does a magnetic field do so?
YesNo

Answers

(a) Yes, an electric field can exert a force on a stationary charged object. A stationary charged object will experience a force in the direction of the electric field due to the Coulombic interaction between the charges.

(b) No, a magnetic field does not exert a force on a stationary charged object. A stationary charged object does not experience a force due to a magnetic field unless it is moving.

(c) Yes, an electric field can exert a force on a moving charged object. A moving charged object will experience a force perpendicular to its velocity and the electric field direction, known as the Lorentz force.

(d) Yes, a magnetic field can exert a force on a moving charged object. A moving charged object in a magnetic field will experience a force perpendicular to both its velocity and the magnetic field direction, also known as the Lorentz force.

(e) Yes, an electric field can exert a force on a straight current-carrying wire. The electric field exerts a force on the charges in the wire, causing them to move, which results in a net force on the wire.

(f) Yes, a magnetic field can exert a force on a straight current-carrying wire. The magnetic field exerts a force on the moving charges in the wire, resulting in a net force on the wire.

(g) Yes, an electric field can exert a force on a beam of moving electrons. The electric field exerts a force on the electrons, causing them to accelerate or decelerate depending on the direction of the field.

(h) Yes, a magnetic field can exert a force on a beam of moving electrons. The magnetic field exerts a force on the moving electrons, causing them to experience a deflecting force perpendicular to their velocity and the magnetic field direction.

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Which term describes the energy an object has due to the motion of its
particles?
A. Magnetic energy
B. Chemical energy
C. Elastic energy
D. Thermal energy

Answers

Answer: The answer is D. Thermal Energy.

Explanation:

Thermal energy is a type of kinetic energy owing to the fact that it results from the movement of particles.

hydroelectric, wind, geothermal, and parabolic solar collection all rely on spinning turbines (connected to a generator) to produce electricity. explain how each provides the force to do so.

Answers

Hydroelectric energy is generated by capturing the energy of flowing water. As water flows through a turbine, the blades of the turbine spin and generate electricity.

How does the different energies provide force?

Wind energy is generated by capturing the kinetic energy of the wind. As wind passes through the turbine, the blades spin and generate electricity.

Geothermal energy is generated by harnessing the natural heat of the Earth’s core. Heat from the Earth’s core is used to generate steam, which is then used to spin a turbine and generate electricity.

Parabolic solar collection is a method of collecting the sun’s energy using large reflective mirrors. The mirrors focus the sunlight onto a central point, which is then used to spin a turbine and generate electricity.

Thus, all of these power sources rely on spinning turbines connected to a generator to produce electricity.

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why can't we fall safely with the help of parachute towards the moon?​

Answers

Answer:

The Moon has no atmosphere so there is no drag on the capsule to slow its descent; parachutes will not work. Lunar landing vehicles were equipped with rocket engines that were fired by the pilot to provide lift — thrust in the opposite direction of descent — during the rapid descent to the Moon's surface.

The moon does not harbor any appreciable atmosphere. Therefore no parachute, no matter how large, will operate properly on the moon. Air is required in order to inflate the parachute and slow down the descending object. Remember geologist Harrison Schmidt, the ONLY scientist to visit the moon? He was one of the last two people to ever touch the lunar surface. (Apollo 17). He demonstrated what would happen when two objects of different masses were dropped simultaneously from about five feet above the moon’s surface. He dropped a hammer and a feather. They fell at the same rate and hit the surface at exactly the same instant! There was no atmosphere to cause the feather to flutter. Note: Careful observers may notice that in videos of the the descending Apollo Lunar Lander (“The Eagle has landed”) lunar dust is kicked up by the craft’s engines. The dust moves out in straight lines, not in billowing clouds! PROOF that the film was made in the airless void of the moon and NOT in some clandestine film studio on Earth. No moon landing hoax!

Masses m1 and m2 are supported by wires that have equal lengths when unstretched. The wire supporting m1 is an aliminum wire 0. 9 mm in diameter, and the one supporting m2 is steel wire 0. 3 mm in diameter. What is the ratio m1/m2 if the two wires stretched by the same amount?

Answers

A wire's ability to elongate (or stretch) under stress is influenced by a number of variables, including the force used, the wire's cross-sectional area, and the material's elastic modulus.

The stiffness or resistance to deformation of a material is measured by the modulus of elasticity, which varies for steel and aluminium.While supporting the masses m1 and m2, let L be the length of each wire when it is not extended, and let L be the common elongation (or stretch) of the wires.

The force exerted on each wire comes from:

F = mg

where g is the gravitational acceleration. The identical amount of stretching is applied to both wires, therefore we have:

F1/A1 = F2/A2

where the cross-sectional areas of the steel and aluminium wires, respectively, are A1 and A2, respectively. A wire of diameter d has a cross-sectional area given by:

A = πd²/4

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Select the correct location on the image.
The image shows the visible light spectrum received from a star. Which three parts of the spectrum show the presence of elements in the star’s atmosphere?

Answers

The visible light spectrum is the range of wavelengths the human eye can detect, ranging from 380 to 700 nanometers.

What are visible light examples?

People think of the sun, light bulbs, candles, and flames when they think of light, but visible light originates from many sources and in many hues. Other visible light sources include television and computer displays, glow sticks, and pyrotechnics.

This is why this area of the electromagnetic spectrum is known as the visible spectrum or colour spectrum. It primarily comprises of seven colours: violet, blue, green, yellow, orange, and red.

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Answer:

It is the three spots where there are lines. Between 400 and 500(the two lines), between 600 and 700(the two lines), and the one line between 700 and 800.

The electric flux through a spherical surface is4.3×104 N⋅m2/C. What is the net charge enclosed by the surface? The net charge enclosed by the surface isμC. The electric flux through a cubical box34 cmon a side is7.5×103 N⋅m2/C. What is the total charge enclosed by the box? The total charge enclosed by the box isμC

Answers

For the electric flux through a spherical surface is 4.3 x 10⁴ N⋅m²/C, then the net charge enclosed by the surface is μC, and for the electric flux through a cubical box 34 cm on a side is 7.5 x 10³ N⋅m²/C, the total charge enclosed by the box is μC.

The electric flux through a spherical surface is 4.3 x 10⁴ N⋅m²/C.

The net charge is Electric Flux = Charge / Surface Area,

so the net charge enclosed is 4.3 x 10⁴ / (4πr²) where r is the radius of the sphere.

Therefore, the net charge enclosed by the surface is μC.

The electric flux through a cubical box 34 cm on a side is 7.5 x 10³ N⋅m²/C.

The total charge is Electric Flux = Charge / Surface Area,

so the total charge enclosed is 7.5 x 10³ / (6a²)

where a is the length of one side of the cube.

Therefore, the total charge enclosed by the box is μC.

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The half life of a radioactive substance is 5 hours. If 5g of the substance is left after 20 hours, determine the original mass of the substance​

Answers

Answer:

The original mass of the substance was 10g.

Explanation:

The half-life of a radioactive substance is the amount of time it takes for half of the substance to decay. In this case, the half-life is 5 hours.

We can use the half-life formula to find the original mass of the substance:

N = N0 * (1/2)^(t/T)

where:

---N0 is the initial mass of the substance

---N is the remaining mass of the substance after time t

---T is the half-life of the substance

We know that after 20 hours, only half of the substance remains:

N = N0 * (1/2)^(20/5) = 0.5 * N0

If we solve for N0, we get:

N0 = N / 0.5 = 5g / 0.5 = 10g

Therefore, the original mass of the substance was 10g.

If the 0. 100-mm diameter tungsten filament in a light bulb is to have a resistance of 0. 200 ω at 20. 0oc , how long should it be?

Answers

The length is 2.78 mm if the 0. 100-mm diameter tungsten filament in a light bulb is to have a resistance of 0. 200 ω at 20 degrees.

The length tungsten filament is 2.78 mm to have a resistance of 0. 200 ω at 20. degrees.

The given data is as follows:

Diameter of tungsten = 0.100 mm

resistance of tungsten = 0.200ω

The resistance (R) of a conductor is calculated by using the formula,

R = ρ × (L/A)

ρ =   resistivity of the material

L =  length of the conductor

A  =  cross-sectional area.

By rearranging the formula to calculate the length,

L = (R × A) / ρ

A = π × r²

A = 3.14 × (5.0 x [tex]10^{-5}[/tex])²

A = 7.85 x [tex]10^{-9}[/tex] m²

The resistivity of tungsten at 20.0°C  =  5.6 x [tex]10^{-8}[/tex] Ωm

L = (0.200 × 7.85 x [tex]10^{-9}[/tex]) / (5.6 x [tex]10^{-8}[/tex])

L = 2.78 x [tex]10^{-3}[/tex] m

L = 2.78 mm

Therefore we can conclude that the length is 2.78 mm to have a resistance of 0. 200 ω at 20 degrees.

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the orbital period of saturn is 29.46 years. determine the distance from the sun to the planet in km

Answers

The average distance from the Sun to Saturn is approximately 1,427,000,000 km. To calculate this, we can use the Third Kepler's Law of Planetary Motion, which states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of the orbit.

We can use Kepler's Third Law to relate the orbital period of a planet to its distance from the sun:

T^2 = (4π^2 / GM) * r^3

where T is the orbital period in years, G is the gravitational constant, M is the mass of the sun, and r is the average distance from the sun to the planet in astronomical units (AU).
Therefore, we can use the formula:

d^3 = (T^2 * 4π^2)/G*M

Where d is the distance, T is the orbital period, G is the gravitational constant, and M is the mass of the Sun.


Plugging in the values:

d^3 = (29.46^2 * 16π^2)/(6.67408 * 1.989 * 10^30)
d = 1,427,000,000 km

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(3)
Four particles are located at points (1,4), (2,3), (3,3), (4,1).?
Find the moments Mx and My and the center of mass of the system, assuming that the particles have equal mass m.
Mx=
My=
xCM=
yCM=
Find the center of mass of the system, assuming the particles have mass 3, 2, 5, and 7, respectively.
xCM=
yCM=

Answers

Given that four particles are located at points (1,4), (2,3), (3,3), (4,1).

The moments Mx and My and the center of mass of the system can be determined as follows:

For equal mass m, the moment Mx is obtained by summing the product of the mass of each particle and the perpendicular distance from the line y=0.

Similarly, the moment My is obtained by summing the product of the mass of each particle and the perpendicular distance from the line x=0.

My = Σ mi*yiMy = (m(1)+m(2)+m(3)+m(4))(4+3+3+1)/4My = 11m

Hence, the moments Mx and My are 10m and 11m, respectively.

For particles with mass 3, 2, 5, and 7 respectively, the x-coordinate and y-coordinate of the center of mass of the system are given by:

xCM = (Σ mixi)/Mx= (3*1+2*2+5*3+7*4)/17= (3+4+15+28)/17= 50/17yCM = (Σ miyi)/My= (3*4+2*3+5*3+7*1)/17= (12+6+15+7)/17= 40/17

Hence, the center of mass of the system is at (50/17, 40/17).

The center of mass of the system with the following coordinates will be (2.76, 2.76). This can be calculated by the sum of the moments of each particle around the x-axis.

What is the center of mass of the system?

Here, we are given four particles that are located at points (1,4), (2,3), (3,3), (4,1). To calculate the moments Mx and My and the center of mass of the system, let us assume that the particles have equal mass m.

Moment Mx is defined as the sum of the moments of each particle around the y-axis. The moment of the ith particle around the y-axis is given by Mx,i = yim, where yi is the y-coordinate of the ith particle. Therefore, the total moment Mx of the system is: Mx = Mx,1 + Mx,2 + Mx,3 + Mx,4 = 4m + 3m + 3m + 1m = 11m

Therefore, Mx = 11m.

Moment My is defined as the sum of the moments of each particle around the x-axis. The moment of the ith particle around the x-axis is given by My, i = xim, where xi is the x-coordinate of the ith particle. Therefore, the total moment My of the system is: My = My,1 + My,2 + My,3 + My,4 = 1m + 2m + 3m + 4m = 10m

Therefore, My = 10m.

The coordinates of the center of mass (xCM, yCM) are given by:

xCM = Σmixi / ΣmiyCM = Σmiyi / Σmi

where, Σmi is the sum of the masses and Σmixi and Σmiyi are the sums of the moments around the y-axis and x-axis, respectively.

If the particles have equal mass m, then Σmi = 4m + 3m + 3m + 1m = 11m.

xCM = (1×4 + 2×3 + 3×3 + 4×1) / 11 = 2.45

yCM = (1×4 + 2×3 + 3×3 + 4×1) / 11 = 2.45

Therefore, the center of mass of the system is (2.45, 2.45).

If the particles have mass 3, 2, 5, and 7, respectively, then Σmi = 3 + 2 + 5 + 7 = 17.

xCM = (1×3 + 2×2 + 3×5 + 4×7) / 17 = 2.76

yCM = (4×3 + 3×2 + 3×5 + 1×7) / 17 = 2.76

Therefore, the center of mass of the system is (2.76, 2.76).

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A straight 2.40 m wire carries a typical household current of 1.50 A (in one direction) at a location where the earth's magnetic field is 0.550 gauss from south to north. *I know there's a lot of questions, but I will rate the you-know-what out of you a) Find the direction of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east. b) Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east. c) Find the direction of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running vertically upward. d) Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running vertically upward. e) Find the direction of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from north to south. f) Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from north to south. g) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

Answers

a) If the current is running from west to east, the force that our planet's magnetic field exerts on this cord is directed upwards
b) The magnitude of the force that our planet's magnetic field exerts on this cord if it is oriented so that the current in it is running from west to east is F =2.64 x 10^-4 N
c) If the current is running vertically upward, the force that our planet's magnetic field exerts on this cord is directed to the left.  west
d) The magnitude of the force that our planet's magnetic field exerts on this cord if it is oriented so that the current in it is running vertically upward is F = 0 zero
e) If the current is running from north to south, the force that our planet's magnetic field exerts on this cord is directed east.
f) The magnitude of the force that our planet's magnetic field exerts on this cord if it is oriented so that the current in it is running from north to south is F = 2.64 x 10^-4 N
g) The magnetic force is not large enough to cause significant effects under normal household conditions.

EXPLANATION

a) The direction of the force that our planet's magnetic field exerts on the cord is perpendicular to both the direction of the current and the direction of the magnetic field, according to the right-hand rule. In this case, if the current is running from west to east, and the magnetic field is from south to north, the force will be directed upwards.

b) The magnitude of the force can be calculated using the formula:

F = BIL sin(theta)

where B is the magnitude of the magnetic field, I is the current, L is the length of the wire, and theta is the angle between the direction of the current and the direction of the magnetic field. In this case, theta is 90 degrees, so sin(theta) = 1. Substituting the given values, we get:

F = (0.550 x 10^-4 T) x (1.50 A) x (2.40 m) x 1

= 2.64 x 10^-4 N

Therefore, the magnitude of the force is 2.64 x 10^-4 N.

c) If the current in the wire is running vertically upward, the force will be directed towards the west.

d) Using the same formula as in part (b), we can calculate the magnitude of the force:

F = (0.550 x 10^-4 T) x (1.50 A) x (2.40 m) x sin(90)

= 0

Therefore, the magnitude of the force is zero.

e) If the current in the wire is running from north to south, the force will be directed towards the east.

f) Using the same formula as in part (b), we can calculate the magnitude of the force:

F = (0.550 x 10^-4 T) x (1.50 A) x (2.40 m) x 1

= 2.64 x 10^-4 N

Therefore, the magnitude of the force is 2.64 x 10^-4 N.

g) The magnitude of the magnetic force in this case is quite small, and under normal household conditions, it is unlikely to cause significant effects. However, in some situations, such as in electrical power transmission systems, the effects of the magnetic force may need to be taken into account.

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as a source of blackbody radiation becomes hotter, the peak in its radiation spectrum moves from the visible to the ultraviolet and beyond. does this imply that the object can no longer be seen by the unaided human eye

Answers

Yes, it is correct that when the source of blackbody radiation becomes hotter, the peak in its radiation spectrum shifts from the visible to the ultraviolet and beyond. Blackbody radiation is electromagnetic radiation emitted from a blackbody or perfect absorber. This is due to the fact that hotter objects emit shorter wavelengths of electromagnetic radiation, which correspond to higher energy photons. Therefore, when an object gets hot enough to emit mostly ultraviolet or X-ray radiation, it will no longer be visible to the unaided human eye because the human eye can only detect radiation within the visible spectrum of about 400 nm (violet) and 700 nm (red). Therefore, a blackbody that emits radiation beyond this range will no longer be seen by the unaided human eye.

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