A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with
the racket for 0.04 s, what is the average force on the ball by the racket?


A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with
the racket for 0.04 s, what is the average force on the ball by the racket?


0.57 N


32 N


98 N


0.00005 N

Answers

Answer 1

From Newton's second law of motion, the average force on the ball by the racket is 98 Newtons. The correct answer is option C

Given that a 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. And the time for contact with the racket is 0.04 s, that is,

mass m = 0.07 kg

velocity v = 56 m/s

time t = 0.04 s

force f = ?

To calculate the average force on the ball by the racket, let us apply Newton's second law of motion.

Impulse = change in momentum

ft = mv

Substitute all the parameters into the equation above

0.04f = 0.07 x 56

make f the subject of the formula

f = 3.92 / 0.04

f = 98 N

Therefore, the average force on the ball by the racket is 98 Newtons. The correct answer is option C

Learn more about momentum here: https://brainly.com/question/7538238

Answer 2

The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

From the question, we are to determine the average force on the ball by the racket.

From the formula,

[tex]F = \frac{mv}{t}[/tex]

Where F is the force

m is the mass

v is the velocity

and t is the time

From the given information

m = 0.07 kg

v = 56 m/s

t = 0.04 s

Putting the parameters into the formula,

we get

[tex]F = \frac{0.07 \times 56}{0.04}[/tex]

[tex]F = \frac{3.92}{0.04}[/tex]

F = 98 N

Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

Learn more on calculating force exerted on an object here: https://brainly.com/question/13590154


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Answer:

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The above question requires a personal answer about your perception of corals, environmental conservation, and the activity you did in the classroom. In that case, I can't answer your question, but I'll show you how to answer it.

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Answer:

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Answer:

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Answer:

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Answers

As the spring is compressed, it performs

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Answer:

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Answers

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Answers

Answer:

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Answer:

Give person above me brainliest

Explanation:

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is ____? ​

Answers

Let, the maximum height covered by projectile be [tex]\sf{H_m}[/tex]

[tex]\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }} [/tex]

Projectile is thrown with a velocity = v Angle of projection = θ

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[tex]\qquad[/tex]______________________________

Then –

[tex]\qquad[/tex] [tex]\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }[/tex]

Now, the vertical component of velocity of projectile at the height half of [tex] \sf{h_m}[/tex] will be –

[tex]\qquad[/tex] [tex]\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }[/tex]

Therefore, the vertical component of velocity of projectile at this height will be–

☀️[tex]\qquad[/tex][tex] \pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }[/tex]

Answer:

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is v sintheta / √2

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Answers

Answer:

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19:10.8

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A 0.24 kg mass with a speed of 0.60 m/s has a head-on collision with a 0.26 kg mass that is traveling in the opposite direction at a speed of 0.20 m/s. Assuming that the collision is perfectly inelastic, what is the final speed of the combined masses?

Answers

Answer:

  0.184 m/s

Explanation:

Momentum is conserved. If the velocity of the 0.24 kg mass is positive, then ...

  m1v1+m2v2 = m3v3

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  v = (0.144 -0.052 kg·m/s)/(0.50 kg) = 0.184 m/s

The speed of the combined masses is 0.184 m/s.

_____

Additional comment

The positive sign indicates the combined masses are moving in the direction of the original 0.24 kg mass.

describe the working principle of a hydraulic press using pascals principle

Answers

Answer:

A hydraulic press works on the principle of Pascal's law, which states that when pressure is applied to a confined fluid, the pressure change occurs throughout the entire fluid. Within the hydraulic press, there is a piston that works as a pump, that provides a modest mechanical force to a small area of the sample.

Explanation:

in any chemical reaction or physical change the mass of the product is ___ The mass of the reactant
a. The relationship cannot be determined by the amount of information given
b. equal to or the same
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Answers

Answer: b. equal to or the same

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Answers

1. • Here, force of gravity on the block = 20 N.

• Therefore, the normal force will also be the same, i.e., 20 N [According to Newton's Third Law, on every action, there is an equal and opposite reaction]

• The coefficient

[tex] u_{k} = 0.4[/tex]

• Force of friction =

[tex]u_{k} \times \: normal \: \: \: force \\ = 0.4 \times 20N \\ = 8N[/tex]

• Hence, the force of sliding friction between the block and the ground is 8 N.

• So, it is option c. 8 N

2. The answer is option d. continue in the same direction with no change in speed.

We know, force = mass × acceleration. When force is 0, then acceleration will also be 0 since mass cannot be 0. So, there will be no change in speed.

3. It is option b. force that is required to give a one kilogram object the acceleration of 1 m/s^2.

Newton is the SI unit of force. As mentioned earlier, force = mass × acceleration. The SI unit of mass and acceleration is Kg and m/s^2 respectively.

So, 1 N = 1 Kg × 1 m/s^2.

4. It is d. not zero.

Acceleration is the change in speed. So, if the force is zero, then acceleration will not occur.

5. Force = 2 N

Acceleration of the object A = 2 m/s^2.

Acceleration of the object B = 1 m/s^2.

Therefore, mass of the object A = 2 N ÷ 2 m/s^2 = 1 Kg

And, mass of the object B = 2 N ÷ 1 m/s^2 = 2 Kg

So, the mass of object B is greater than that of object A.

Hence, the answer is option c. Object B has more mass.

Hope you could get an idea from here.

Doubt clarification - use comment section.

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