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ZanzabumX [31]
2 years ago
11

A 300N box on a 43 degree angle.

Physics
1 answer:
Ksenya-84 [330]2 years ago
7 0

Answer:is this a question??? I’m so confused

Explanation:

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1500 kg wrecking ball traveling at a speed of 3.5 m/s hits a wall that does not crumble but is pushed back 75 cm. If the wreckin
Rudiy27

Answer:

The size of the force that pushes the wall is 12,250 N.

Explanation:

Given;

mass of the wrecking ball, m = 1500 kg

speed of the wrecking ball, v = 3.5 m/s

distance the ball moved the wall, d = 75 cm = 0.75 m

Apply the principle of work-energy theorem;

Kinetic energy of the wrecking ball = work done by the ball on the wall

¹/₂mv² = F x d

where;

F is the size of the force that pushes the wall

¹/₂mv² = F x d

¹/₂ x 1500 x 3.5² = F x 0.75

9187.5 = 0.75F

F = 9187.5 / 0.75

F = 12,250 N

Therefore, the size of the force that pushes the wall is 12,250 N.

7 0
2 years ago
What is the magnitude b of the magnetic field at the location of the charge due to the current-carrying wire?
elena-14-01-66 [18.8K]

Answer:

k

Explanation:

3 0
3 years ago
1)Atmospheric decreases with the increase with in height?​
lina2011 [118]

Answer:

Explanation:

pressure decreases with increasing altitude. The pressure at any level in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels.

5 0
2 years ago
Name the physical quantity measured by the slope of the velocity time graph for uniform motion
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Acceleration. If the line is linear on a velocity time graph, the acceleration does not change, meaning its 0
8 0
3 years ago
1. A 14-cm tall object is placed 26 cm from a converging lens that has a focal length of 13 cm.
AURORKA [14]

Answer:

a) Please find attached the required drawing of light passing through the lens

By the use of similar triangles;

The image distance from the lens = 26 cm

The height of the image = 14 cm

c) The image distance from the lens = 26 cm

The height of the image = 14 cm

Explanation:

Question;

a) Determine the image distance and the height of the image

b) Calculate the image position and height

The given parameters are;

The height of the object, h = 14 cm

The distance of the object from the mirror, u = 26 cm

The focal length of the mirror, f = 13 cm

The location of the object = 2 × The focal length

Therefore, given that the center of curvature ≈ 2 × The focal length, we have;

The location of the object ≈ The center of curvature of the lens

The diagram of the object, lens and image created with MS Visio is attached

From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m

The height of the image = The height of the object - 14 cm

b) The lens equation is used for finding the image distance from the lens as follows;

\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}

Where;

v = The image distance from the lens

We get;

v = \dfrac{u \times f}{u - f}

Therefore;

v = \dfrac{26 \times 13}{26 - 13} = \dfrac{26 \times 13}{13} = 26

The distance of the image from the lens, v = 26 cm

The magnification, M =v/u

∴ M = 26/26 = 1, therefore, the object and the image are the same size

Therefore;

The height of the image = The height of the object = 14 cm.

5 0
2 years ago
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