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3 years ago

What is the momentum of a dog of mass 20 kg that is running east with a speed of 3 m/s?

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

4 0

Answer:

60 kg.m/s

Explanation:

Momentum is the product of mass and velocity

p=m*v where m is mass in kg and v is velocity in m/s

Given that;

m=20kg and v=3m/s then;

p= 20×3= 60kgm/s

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4. Compare how thermal and electrical energy transfer between objects and give an example of how you can

Leokris [45]

Answer:

4) we take a can of cold soda there is a transfer from hand to can. a current passes through a light bulb, it becomes incandescent and gives ligh

5) the law of energy conservation states that energy is neither created nor destroyed, therefore what happens is that you use energy and transform it into another type of energy,

Explanation:

4) This exercise is asked to give some examples of thermal and electrical energy transfer

the transfer of thermal energy occurs when we grasp a hot or cold body due to the difference in temperature between the body and us there is an exchange of heat, for example when drinking a cup of hot coffee there is a transfer of energy from the cup to the hand .

If we take a can of cold soda there is a transfer from hand to can.

Electrical transfer occurs, for example, when a current passes through a light bulb, it becomes incandescent and gives light.

In all modern electrical appliances there is transformation of electrical energy

5) the law of energy conservation states that energy is neither created nor destroyed, therefore what happens is that you use energy and transform it into another type of energy,

For example, when you lift a box, the potential energy of the box increases with height, but the energy of the person decreases because it does work that is negative.

In amusement park rides, the energy accumulated in one part of the game generally with height is transformed into energy of movement in another part of the game, but the total energy remains constant.

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3 years ago

Which of Newton's laws explains why satellites need very little fuel to stay in oribit?

Angelina_Jolie [31]

Sattelites don't need any fuel to stay in orbit. The applicable law is...."objects in motion tend to stay in motion". Having reached orbital velocity, any such object is essentially "falling" around the earth. Since there is no (or at least very little) friction in the vacuum of space, the object does not slow.... It simply continues.

Sattelites in "low" earth orbit do encounter some friction from the very thin upper atmosphere, and they will eventually "decay".

4 0

3 years ago

Which of the following criteria can help Linda classify gases as greenhouse gases? gas molecules having three or more atoms gas

ratelena [41]

I think the correct answer would be the third option. The criteria that could help Linda in classifying whether the gases are greenhouse gases would be gas molecules having at least one oxygen atom. Most of the greenhouse gases has an oxygen atom in their structures especially those that naturally occurs. These gases are CO2, H2O vapor and nitrous oxide.

7 0

3 years ago

Read 2 more answers

Why are the coldest places on earth found near the poles

Nata [24]

Polar regions do not receive direct sunlight during the winter months due to the tilt in the Earth's axis. Hence, polar regions can get very cold. Antarctica is the coldest place on Earth. The coldest places on Earth tend to be located near the poles. Hope this answers the question.

6 0

3 years ago

Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their

geniusboy [140]

Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

Explanation:

In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

$m_{1} v_{x1}+m_{2}v_{x2}=(m_{1}+m_{2})v_{fx}\\m_{1} v_{y1}+m_{2}v_{y2} =(m_{1}+m_{2})v_{fy}$

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

$1*13 + 4*0 = (1+4) v_{fx}\\v_{fx}=\frac{13}{5} =2.6\\1*0 + 4*13 = (1+4) v_{fy}\\v_{fy}=\frac{13*4}{5} =10.4\\$

The magnitude of the final velocity of the wreck can be found as:

$v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72$

The final velocity has an intensity of roughly 11 m/s

As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

$\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}$

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.

5 0

3 years ago