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dexar [7]
3 years ago
13

A Nerf gun is shot straight up from the ground with an initial velocity of 14.0 m/s. What is the total time the projectile is in

the air?
PLEASE SHOW WORK WILL MARK BRAINLIEST
Physics
1 answer:
Sergeeva-Olga [200]3 years ago
7 0

The projectile has a height <em>h</em> at time <em>t</em> given by

<em>h</em> = (14.0 m/s) <em>t</em> - 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for <em>t</em> when <em>h</em> = 0 :

0 = (14.0 m/s) <em>t</em> - 1/2 <em>g t</em> ²

0 = 1/2 <em>t</em> (28.0 m/s - <em>g t</em> )

1/2 <em>t</em> = 0   <u>or</u>   28.0 m/s - <em>g</em> <em>t</em> = 0

The first equation says <em>t</em> = 0, which refers to the moment the gun is first fired, so we ignore that solution. We're left with

28.0 m/s - <em>g t</em> = 0

<em>t</em> = (28.0 m/s) / <em>g</em>

<em>t</em> = (28.0 m/s) / (9.80 m/s²)

<em>t</em> ≈ 2.86 s

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Is gravity a non-contact form?
sattari [20]
No, we cannot touch gravity nor can we physically see it. We can only see how it works.
8 0
3 years ago
If jack was traveling north 120 miles and it took him 3 hours to get there. what is the velocity that jack was traveling? is thi
Dimas [21]

Answers:

40 mp/h; Vector

Reason:

120/3 is 40 miles per hour.

Velocity is a vector measurement.

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- Amanda

4 0
3 years ago
A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the averag
andrey2020 [161]

Answer:

The small car and the truck experience the same average force.

Explanation:

Here we need to remember two of Newton's laws.

The second one says that:

F = m*a

force equals mass times acceleration.

And the third one says that;

"If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A"

From the third law, if the car experiences a force F due to the impact with the truck, then the truck experiences the same force F due to the impact.

But this seems odd, because we would expect to see the car being more affected by the impact, right?

Well, this is explained by the second law.

Suppose that the mass of the car is m, and the mass of the truck is M.

such that M > m

Then for the small car we have:

F = m*a

And for the truck:

F = M*a'

Because the force is the same for both of them, we can write:

m*a = M*a'

a = (M/m)*a'

because M > m, then M/m > 1.

This means that the acceleration that the car experiences is larger than the acceleration for the truck, and this is why we would see that the car seems more affected by the impact, regardless of the fact that both vehicles experience the same force in the impact.

6 0
3 years ago
2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or
krek1111 [17]

a) The work done by the tree is -1.44\cdot 10^5 J

b) The amount of force applied is 2880 N

Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

W=Fd

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

W=-1.44\cdot 10^5 J is the work done

d=50.0 cm = 0.50 m is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0
3 years ago
How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?
Illusion [34]
Power is defined as the rate at which the body is doing work:
P=\frac{W}{t}
Work is defined as displacement done by the force times that displacement:
W=F\cdot h
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
W=62N\cdot10m=620J
of work.
Now we just divide that by 5s to get how much power is required:
P=\frac{620J}{5s}=124W
5 0
3 years ago
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