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2 years ago

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

2 answers:

5 0

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: brainly.com/question/7538238

Naddika [18.5K]2 years ago

4 0

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,

$F = \frac{mv}{t}$

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

$F = \frac{0.07 \times 56}{0.04}$

$F = \frac{3.92}{0.04}$

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: brainly.com/question/13590154

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How much force is needed to keep the bowling ball moving towards the pins once it has

vazorg [7]

Answer:

Explanation:

The amount of force needed needs to be greater than all the forces acting in the opposite direction that the bowling ball was thrown. This includes air resistance, floor friction, gravity, and any other force involved. As long as the force acting on the bowling ball that is causing it to go in the direction of the pins is slightly greater than the opposite acting forces then it will continue in that direction. Since no values are provided we cannot calculate the actual precise value of force needed.

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2 years ago

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iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):

$\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}$

$m = \frac{1}{2}\cdot M$

And the resulting equation is:

$I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)$

$I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}$

$I = \frac{15}{32}\cdot M\cdot R^{2}$

The moment of inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

5 0

2 years ago

Un tubo cilindrico hueco de cobre mide 3 m de longitud tienen un diametro exterior de 4cm y un diametro interior de 2 cm¿cuanto

Irina-Kira [14]

Answer:

W = 9.93 10² N

Explanation:

To solve this exercise we must use the concept of density

ρ = m / V

the tabulated density of copper is rho = 8966 kg / m³

let's find the volume of the cylindrical tube

V = A L

V = π (R_ext ² - R_int ²) L

let's calculate

V = π (4² - 2²) 10⁻⁴ 3

V = 1.13 10⁻² m³

m = ρ V

m = 8966 1.13 10⁻²

m = 1.01 10² kg

the weight of the tube

W = mg

W = 1.01 10² 9.8

W = 9.93 10² N

4 0

2 years ago

A balloon is filled to a volume of 7.00*10^2 mL at a temperature of 20.0°C. The balloon is then cooled at constant pressure to a

wolverine [178]

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

$\frac{V}{T}=const.$