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tekilochka [14]
3 years ago
12

A car moves with a velocity of 20 m/s to east and the other with velocity of 15 m/s to west. If they move from the same point, h

ow far will each of them travel in 2 minutes? Calculate the distance between them after 2 minutes​
Physics
1 answer:
Deffense [45]3 years ago
3 0

Answer:

4.2 km

Explanation:

CAR A(velocity =20m/s to East)

In 1s, it covers 20m.

In 120s(2min),it covers20*120m.=2400m=2.4 km

CAR B(velocity =15m/s to west)

In 1s, it covers 15m.

In 120s, it covers 15*120m.=1800m=1.8km

Now, distance between the two cars after 2 minutes =2.4 km+1.8 km=4.2km

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Olympic gold medalist Michael Johnson runs one time around the track 400 meters in 38 seconds what is his displacement what is h
Sveta_85 [38]
Displacement = 0, assuming that he runs back to original position
Average velocity is displacement/ time, since displacement =0, velocity is also 0
8 0
3 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
What would the person’s mass be on the earth? Part B pls
Dafna1 [17]

Answer:

I hopes it helps

thank you

4 0
3 years ago
What is the terminal velocity of a penny falls off a skyscraper
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The penny will reach terminal velocity at 50 ft. Then it will travel 25 mph till it reaches ground.
7 0
4 years ago
Use the law of conservation of energy (assume no friction nor air resistance) to determine the kinetic and potential energy at t
slava [35]

Answer:

Part A

1) At the starting point, we have;

PE = 40,000 J

2) PE = 0 J, KE = 40,000 J

3) KE = 20,000 J

4) PE = 15,000 J

5) KE = 32,500 J

6) KE = 40,000 J, PE = 0 J

7) KE = 35,000 J

8) KE = 40,000 J, PE = 0 J

Part B

The total Mechanical Energy = ME = 40,000 J

At the final point, we have;

ME = KE + PE = 40,000 J + 0 J = 40,000 J

Explanation:

Part A

By the law of conservation of energy, we have;

ME = PE + KE

Where;

ME = The total Mechanical Energy of the system

PE = The Potential Energy of the system

KE = The Kinetic Energy of the system

Where there is no friction, we have;

At the final stage, KE = 40,000 J. PE = 0 J

Therefore, ME = PE + KE = 40,000 J + 0 J = 40,000 J

1) At the starting point, we have;

KE = 0 J, therefore, PE = ME - KE = 40,000 J - 0 J = 40,000 J

2) At the bottom of the roller coaster, at the same level the PE is taken as PE = 0 J at the final stage, we have;

PE = 0 J, therefore, KE = ME - PE = 40,000 J - 0 J = 40,000 J

3) Where PE = 20,000 J, KE = ME - PE = 40,000 J - 20,000 J = 20,000 J

4) Where KE = 25,000 J, PE = ME - KE = 40,000 J - 25,000 J = 15,000 J

5) Where PE = 7,500 J, KE = ME - PE = 40,000 J - 7,500 J = 32,500 J

6) At the bottom KE = 40,000 J, PE = 0 J

7) Where PE = 5,000 J, KE = ME - PE = 40,000 J - 5,000 J = 35,000 J

8) KE = 40,000 J, PE = 0 J

Part B

The given that there is no friction nor air resistance, the total Mechanical Energy, ME, is constant and equal to the sum of the Potential Energy, PE and the Kinetic Energy, KE, as follows;

ME = KE + PE

At the final point, we have;

ME = 40,000 J + 0 J = 40,000 J

The total Mechanical Energy = ME = 40,000 J

8 0
3 years ago
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