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inessss [21]
2 years ago
15

A tennis player swings and hits the ball away. How does the force of the tennis racket affect the motion of the ball

Physics
1 answer:
defon2 years ago
4 0

The force of the racket affects the ball's motion because it changes the momentum of the ball.

<h3>Impulse received by the ball</h3>

The impulse received by the ball through the racket affects the motion because it changes the momentum of the ball.

The ball which is initially at rest, will gain momentum after been hit with the racket.

J = ΔP = Ft

where;

  • J is the impulse received by the ball
  • ΔP is change in momentum of the ball
  • F is the applied force
  • t is the time of action

Thus, the force of the racket affects the ball's motion because it changes the momentum of the ball.

Learn more about impulse here: brainly.com/question/25700778

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The gravitational force between two objects is ______ proportional to the products of the masses and _______ proportional to the
Rudiy27

directly, inversely ... newton's grav eqn

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3 years ago
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there are 1.6 km in a mile. the distance between two cities is 248 miles. How many kilometers apart are the two cities?
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3 0
3 years ago
Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is
victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = \frac{-kQq}{r^2}

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = \frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2} = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive  x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = \frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2} = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

3 0
3 years ago
200 kW of solar radiation is shining on a 300 m^2 parking lot. What is the insulation on the parking lot?
ale4655 [162]

That's "<em><u>insolation</u></em>" ... not "insulation".

'Insolation' is simply the intensity of solar radiation over some area.

If 200 kW of radiation is shining on 300 m² of area, then the insolation is

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Note that this is the intensity of the <em><u>incident</u></em> radiation.  It doesn't say anything
about how much soaks in or how much bounces off.

Wait ! 
I just looked back at the choices, and realized that I didn't answer the question
at all.  I have no idea what  "1 sun"  means.  Forgive me.  I have stolen your
points, and I am filled with remorse.

Wait again !
I found it, through literally several seconds of online research.

           1 sun = 1 kW/m².

So 2/3 of a kW per m²  =  2/3 of 1 sun

That's between 0.5 sun and 1.0 sun.

I feel better now, and plus, I learned something.


7 0
3 years ago
Carlos wanted to know how positive and
Iteru [2.4K]

Answer:

it Give only one of them a positive or negative charge

5 0
2 years ago
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