Find the velocities of iron blocks weighing 50 kg and 100 kg if they are dropped from a height of 100 m at the same time.

Answers

Answer 1

Answer:

44.294 m/s

you might want to double check the answer depending on what you use for g. some people use 9.81, some 9.8, and some 10. that can effect the final answer

Explanation:

use the formula [tex]v^{2}-v_{0}^{2} = 2ax[/tex], which can be moved around to get [tex]v = \sqrt{2ax}[/tex]

note that mass does not effect the velocity

since the blocks were dropped from rest, v_0 = 0m/s. a = g = 9.81 m/s^2, and x = h = 100m.

plug in the values for the variables to get v = sqrt(2(9.81)(100)) = 44.294 m/s for both blocks


Related Questions

What other transformation is used to create this pattern

Answers

The other transformation used in creating this pattern other than translation is 180° rotation.

What is pattern transformation?

Pattern transformation refers to the process of altering or manipulating a basic pattern in order to create a new and unique design. This can involve changing the scale, orientation, color, or other characteristics of the original pattern.

One common method of pattern transformation is through repetition, where a pattern is repeated in a regular or irregular arrangement. This can create a sense of movement, rhythm, or complexity within the design.

The pattern in the image contains translation and rotation (180°) symmetries, the pattern is referred to as SPINNING HOP according to Conway.

Learn about Pattern transformation here https://brainly.com/question/28040796

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A 0.60-kg block initially at rest on a frictionless, horizontal surface is acted upon by a force of 7.0 N for a distance of 2.0 m. How much farther would the force have to act for the block to have 57 J of kinetic energy?

Answers

Answer:

Explanation:

o have 57 J of kinetic energy?

We can use the work-energy principle to solve this problem. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. The work done by a constant force on an object is given by the product of the force and the distance over which the force is applied.

Let's first find the initial kinetic energy of the block, which is zero because the block is initially at rest. Then we can find the work done by the force:

W = Fd = (7.0 N)(2.0 m) = 14 J

The work done by the force is 14 J. We want to find the additional distance the force would have to act to give the block a total kinetic energy of 57 J. Let x be the additional distance:

Work done by force over x distance:

W = Fd = (7.0 N)(x) = 7x J

The total work done on the block is the sum of the work done by the force and the change in kinetic energy:

W_total = W + ΔK

where ΔK is the change in kinetic energy.

At the final position, the block has 57 J of kinetic energy, so:

W_total = 57 J

We can now solve for the additional distance x:

W_total = W + ΔK

57 J = 14 J + (1/2)mv_f^2

where v_f is the final velocity of the block.

Since the block starts from rest, the final velocity is given by:

v_f^2 = 2ΔK / m

v_f^2 = 2(57 J) / 0.60 kg = 95 m^2/s^2

v_f = sqrt(95) = 9.746 m/s

Now we can solve for x:

57 J = 14 J + (1/2)(0.60 kg)(9.746 m/s)^2 - (1/2)(0.60 kg)(0 m/s)^2

57 J = 14 J + 27.8 J + 0

57 J - 14 J - 27.8 J = 7x J

15.2 J = 7x J

x = 2.17 m

Therefore, the force would have to act for an additional distance of 2.17 m for the block to have 57 J of kinetic energy.

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