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SVEN [57.7K]
3 years ago
9

The Hulk lifts a train into the air. He applies a force of 4000 N and lifts the train 10 m. How much work does the Hulk do on th

e train?
Physics
1 answer:
Sphinxa [80]3 years ago
4 0

Answer:

hulk smash

Explanation:

hulk smash

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Please answer this. Science 7th grade.
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I think the answer should be a I’m not sure tho
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3 years ago
Which of these statements is most likely correct about the Nebular theory of formation of planets?
exis [7]
Correct answer is D.

Explanation:
A) is not correct answer because this type of theory can not become law. Laws are properties that are same in any part of universe. Nebular theory is not correct for every part of universe.

B) is not correct answer because this theory could be replaced if some evidence show that some other theory is more likely to be correct.

C) is not correct answer because the study has been done on other nebulas in our galaxy. There are many nebulas and by obserwing them this theory was developed.
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4 years ago
Catherine gently pushes the tip of her finger against the eraser of her pencil and the pencil does not move. Which of the follow
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The best illustration that represents the interaction is D
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3 years ago
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7. A force of 100 N acting on a body gives it a speed of 200 m/s in 2
alekssr [168]

Answer:

Choice a. 1 kg, assuming that all other forces on the object (if any) are balanced.

Explanation:

By Newton's Second Law,

\displaystyle a = \frac{\Sigma F}{m},

where

  • a is the acceleration of the object in \text{m}\cdot\text{s}^{-2},
  • \Sigma F is the net force on the object in Newtons, and
  • m is the mass of the object in kilograms.

As a result,

\displaystyle m = \frac{\Sigma F}{a}.

Assume that all other forces on this object are balanced. The net force on the object will be 100\;\text{N}. The net force is constant. Acceleration should also be constant and the same as the average acceleration in the two seconds.

<h3>What is the average acceleration of this object?</h3>

\displaystyle \begin{aligned}\text{Acceleration} &= \text{Average Acceleration}=\frac{\text{Change in Velocity}}{\text{Time Taken}}\end{aligned}.

\displaystyle {a} = \frac{200\;\text{m}\cdot\text{s}^{-1}}{2\;\text{s}}=100\;\text{m}\cdot\text{s}^{-2}.

<h3>Apply Newton's Second Law to find the mass of the object.</h3>

\displaystyle m = \frac{\Sigma F}{a} = \frac{100\;\text{N}}{100\;\text{m}\cdot\text{s}^{-2}} = 1\;\text{kg}.

6 0
4 years ago
Read 2 more answers
When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some poi
eimsori [14]

Answer: The force of kinetic friction is smaller than that of static friction, but F_g  remains the same.

Explanation:

The situation is same as when a book is pushed with an increasing force on a table; When the force is low, book doesn't move, until that under a given force starts moving, and then it goes on movement even if the force decreases a bit.

The physical explanation for this, that friction force adopts any value needed to avoid to move the object, till a limit value is achieved, called static friction force, equal to the normal force times the static friction coefficient.

Once in movement, the kinetic friction coefficient replaces the static one , and  in general is lower than the static one, so the force diminishes.

In the case of the box sliding down the board, the force that tries to move the object down the board, is the component of the weight parallel to the board, that can be showed that being equal to the weight times the sinus of the angle of the board with the horizontal, as follows:

F_g = m g sin θ

When θ increases, F_g does the same, so friction force always has the same magnitude than F_g (but opposite direction) so the box doesn't move, till that θ takes a value that produces a F_g equal to static friction force.

Beyond this limit angle, F_g (remaining the same for a given angle) is greater than the kinetic friction force, and the box slides.

In the limit, when θ=90º, sin θ =1⇒ F_g = mg, so the object is in free fall.

6 0
4 years ago
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