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myrzilka [38]
3 years ago
11

How much work is done when a 1-kg mass is raised a vertical distance of 1 m

Physics
1 answer:
makkiz [27]3 years ago
5 0
Weight = mass times gravity. . . . Work = (weight) x (height) = (mass) x (gravity) x (height) = (1) x (9.8) x (1) = 9.8 joules.
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A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the
liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0
3 years ago
A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result
Over [174]
B) law of conservation of momentum

It states that the total momentum of a system before impact is the same as the total momentum of the system after impact.

In this case total momentum before impact:

10kg*5m/s  + 5kg * 0m/s = 50 kg m/s

After Impact:

10kg*0m/s + 5kg*10m/s = 50 kg m/s

You can see the momentum before and after impact is same as 50 kg m/s  

Of course we assumed that the first cart stopped after the impact, and there are no energy losses.
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What is the basic difference between a nuclear fission reaction and a nuclear fusion reaction?
Ipatiy [6.2K]
The answer is a. nuclear fission is breaking up
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How does charging by conduction work?
anzhelika [568]

Answer:

During charging by conduction, both objects acquire the same type of charge. If a negative object is used to charge a neutral object, then both objects become charged negatively. ... In this case, electrons are transferred from the neutral object to the positively charged rod and the sphere becomes charged positively.

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Can a small child play with a fat child on a see saw? explain how?
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Closer u get 2 the center the more balanced out your weight will b
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