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Sati [7]
2 years ago
14

A football has a mass of 2 kg and it accelerates at 20 m/s2. What is its force? (You can use a calculator)

Physics
2 answers:
attashe74 [19]2 years ago
8 0
10 mps is the answer
kap26 [50]2 years ago
6 0

Answer:

The answer is 40mps.

Explanation:

1. Force = (mass of an object) x (acceleration)

2. Force = ( "has a mass of 2 kg" ) x ( "accelerates at 20 m/s^2" )

3. Plug the numbers into the formula

4. F = 2 x 20

5. F = 40 mps

<em>Hope I helped!! Please mark me branliest :)</em>

- sophi

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A.  It is zero.

Explanation:

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2 years ago
A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w
professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

\theta=70\° The angle at which the ball leaves the bat

V_{o}=55 m/s The initial velocity of the ball

g=-9.8 m/s^{2} The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: x

Firstly we need to know this velocity has two components:

<u>Horizontally:</u>

V_{ox}=V_{o}cos \theta (1)

V_{ox}=55 m/s cos(70\°)=18.811 m/s (2)

<u>Vertically:</u>

V_{oy}=V_{o}sin \theta (3)

V_{oy}=55 m/s sin(70\°)=51.683 m/s (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when V=0 and the time t is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find t:

V=V_{o}+gt=0 (5)

Isolating t:

t=\frac{-V_{o}}{g} (6)

t=\frac{-55 m/s}{-9.8 m/s^{2}} (7)

t=5.61 s (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time T it takes the complete parabolic path, which is twice t:

T=2t=2(5.61 s)=11.22 s (9)

With this result in mind, we can finally calculate how far the ball travels in the air:

x=V_{ox}T (10)

Substituting (2) and (9) in (10):

x=(18.811 m/s)(11.22 s) (11)

Finally:

x=211.059 m

8 0
3 years ago
Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin
Sergio [31]

Answer:

Explanation:

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= - GMm/ R

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Potential difference

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= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed  to launch an object from the surface of Earth to a height h above the surface.

Extra  energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / ( R + h)

8 0
3 years ago
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