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3 years ago

A runner has an original velocity of 6 m/s and slows to a final velocity of 0 m/s. If the runner covers a

Physics

1 answer:

myrzilka [38]3 years ago

7 0

Answer:

4 s

Explanation:

Given:

Δx = 12 m

v₀ = 6 m/s

v = 0 m/s

Find: t

Δx = ½ (v + v₀) t

12 m = ½ (0 m/s + 6 m/s) t

t = 4 s

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Imagine a box sitting on a shelf. What forces are acting on the box?

romanna [79]

In case of an object sitting at rest on another base, there are two equal and opposite forces – Normal force and the gravity.

Answer: Option A

<u>Explanation: </u>

When an object is placed at rest position on another object, there is a force exerted by the surfaces of the two contact objects. This force is denoted as Normal Force.

When an object such as a box is placed on a shelf, its surface exerts a contact force on the base of the shelf- The Normal force directed upward. Meanwhile, the gravity stays at its action and tries to pull the box towards itself.

Both of these forces however are equal and opposite and therefore, there is zero net force on the box. That's why it remains at rest, holding on Newton's third law.

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3 years ago

a thunderclap sends a sound wave through the air and the ocean below The thunderclap sound wave has a constant frequency of 50 H

Montano1993 [528]

Answer:

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Explanation:

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3 years ago

When a trebuchet counterweight is hoisted by soldiers, what form of energy is

vladimir1956 [14]

10. A safe place to use the trebuchet would be away from other buildings and people. A good example of a place would be a large field with no nearby structures.
14. Many factors need to be kept consistent throughout the experiment. One example of a variable that would need to be consistent is the weight and size of the projectile.
15. It is important to do many trials so that you can make sure that the results of each trial are nearly the same. If they are all vastly different, then it means that something has gone wrong.

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3 years ago

Read 2 more answers

20.0 -kg cannonball is fired from a cannon with muzzle speed of 1000m/s at an angle of 37.0° with the horizontal. A second ball

Dovator [93]

The mechanical energy for the first and the second ball is

$10 ^{7} \: joules.$

Mass of the first ball = 20 kg

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 37°

The maximum height reached by the first ball is,

$= \frac{ u {}^{2} _{1}sin {}^{2} θ}{2g}$

$= \frac{ {1000}^{2} sin {}^{2}37°}{2 \times 9.8}$

$= 18478.69 \: m$

The maximum height of the first cannonball is 17478.69 m.

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 90 °

$= \frac{ u {}^{2} _{2}sin {}^{2} θ}{2g}$

$= \frac{ 1000{}^{2}sin^{2} 90°}{2 \times 9.8}[tex] = 51020.41 \: m$

For the first ball, total mechanical energy= Potential energy at maximum height + kinetic energy at the maximum height

So, the total mechanical energy is,

= mgh \: + \frac{1}{2}mv {}^{2} _{x}[/tex]

$= 20 \times 9.8 \times 18478.64 \times \frac{20}{2} (1000 \: cos37 °)$

$= 10 ^{7} The potential energy at the maximum height, = m _{2}gh$

$= 20 \times 9.8 \times 51020.41$

$= 10 ^{7} \:J$

Therefore, the total mechanical energy for the first and the

$\:second \: cannonball \: is \: 10 ^{7} \:joules.$

To know about energy, refer to the below link:

brainly.com/question/1932868

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5 0

2 years ago

An object is five focal lengths from a concave mirror.how do the object and image heights compare?

enot [183]

An object distance is presented as s = 5f and we know that the mirror equation relates the image distance to the object distance and the focal length.

The mirror equation is 1/f = 1/s + 1/s’ where the variable f stands for the focal length of the mirror. Variable (s) represents the distance between the mirror surface and the object and the variable <span>(s’) represents the distance between the mirror surface and the image. </span>

In addition, a concave mirror will have a positive focal length (f) and a convex mirror will have a negative focal length (f).

Now, we then have 1/f = 1/5f + 1/s’ which is s’ = 5f/4

Then we get the magnification ratio that expresses the size or amount of magnification or reduction of the object or image and to get the magnification, we use this equation: M= s’/s

M= 5f/4x5f

s’ = 1/4s

Therefore, the image height is one fourth of the object height

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3 years ago