All categories

3 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]3 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

You might be interested in

A block weighing 30kg is moved at a constant speed over a horizontal surface by a force of 100 N applied parallel

DENIUS [597]

30-100=70^3•É+ 30 =3^E

5 0

3 years ago

A maintenance worker wants to torque an engine bolt to 65.0 N m. If the torque wrench is 35cm in length, what is force applied t

harina [27]

Answer:

Force = 186 N

Explanation:

Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :

$Torque = \vec{r} \times \vec{F}$

Where $\vec{r}$ is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), $\vec{F}$ is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench. By doing this we have that $r$ (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

$35cm * \dfrac{1m}{100cm} = 0.35m$

Now using the torque formula:

$Torque = \vec{r} \times \vec{F}$

$Torque = rFsin(\theta)}$

Where $\theta$ is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector $\theta = 90°$ , thus :

$Torque = rFsin(\theta)}$

$65 N m = (0.35m)Fsin(90°)}$

$F = \dfrac{65N m} {(0.35m)sin(90°)}$

$F = \dfrac{1300} {7}N$

so the force is approximately 186 N.

4 0

4 years ago

A rubber bullet of mass m=0.025\,\mathrm{kg}m=0.025kg traveling at velocity v_0 = 50\,\mathrm{m/s}v 0 =50m/s hits an iron bloc

Pavel [41]

Answer: 0.001 m

Explanation:

In order to get the máximum height reached by the iron block, we can use the energy conservation principle, as all the kinetic energy impressed upon the iron blck by the rubber bullet during the collision, becomes gravitational potential energy, as follows:

½ m v2 = m. g. h (1)

We don´t know the value of v, but if we look to the collision, and we asume no external forces act during it, the total momentum must be conserved.

The initial momentum, as the block is at rest, is just the one due to the rubber bullet:

P1 = mb . vb = 0.025 kg. 50 m/s = 1.25 kg. m/s

The final momemtum, is just the sum of the one due to the bullet (after being bounced back) and the one for the iron block:

P2 = mb . vfb + mib . vib= 0.025 Kg. (-35 m/s) + 15 Kg.vib

As we have already said, P1 = P2, so we can write the following equation:

0.025 Kg. (-35 m/s) + 15 Kg. vib = 1.25 Kg. m/s.

Solving for vib, we have:

vib = 0.14 m/s

Now, we can replace this value in the equation (1) above:

½ . 15 Kg. (0.14)2 (m/s)2 = 15 Kg. 9.8 m/s2. H

Solving for H, we have:

H = 0.001 m

3 0

3 years ago

If one 9V battery is used in a circuit with a total resistance of 39Ω, what is the current in the circuit?

Ket [755]

Answer:

Using V= IR

I= 0.2307 Ampere

6 0

3 years ago

As the distance between two charged objects increases, the strength of the electrical force between the objects

AveGali [126]

As the distance between two charged objects increases, the strength of the electrical force between the objects <em>decreases</em>.

7 0

3 years ago