Ask question

Ask question

All categories

2 years ago

14

When a 0.058 kg tennis ball is served, it accelerates from rest to a speed of 45.0 m/s. The impact with the racket gives the bal

l a constant acceleration over a distance of 0.440 m. What is the magnitude of the net force acting on the ball?
a.What is the average velocity of the ball?

b.Over what time period was the ball struck?

c. What is the acceleration of the ball?

d.What is the net force on the ball?

e.If the ball had been heavier, but experienced the same change in velocity, would the applied force have to be greater or lesser than before?

1 answer:

AURORKA [14]2 years ago

4 0

Answer:

Explanation:

Note : the questions do not come in the right order.

Data:

Mass = 0.0518kg

Velocity = 45 m / s

Distance (s) = 0.44m

C) what's the acceleration on the ball?

Using equation of motion,

V² = u² + 2as

V² - u² = 2as

a = (v² - u²) / 2s

a = (45² - 0) / 2 * 0.44 [the ball was at rest]

a = 2025 / 0.88

a = 2301.136m/s²

D) The net force on the ball?

Force = mass * acceleration

F = m*a

F = 0.0518 * 2301.136

F = 119.199N

The force acting on the ball was 133.465N

F = 133.47N

b) time period the ball was struck.

From the relationship between impulse and momentum,

Ft = m * v

133.47 * t = 0.058 * 45

t = 2.61 / 133.47

t = 0.01955s

a) average velocity (V) = total distance covered / total time taken

V = s / t

V = 0.44 / 0.01955

V = 22.50m/s.

e) if the ball was heavier and still experienced the same velocity, the applied would've been lesser than before.

You might be interested in

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

The Coulomb constant is $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

The charge on the electron/proton is $e = 1.6*10^{-19} \ C$

The mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

The mass of electron is $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

$m_{electron} * g = \frac{ k * e^2 }{r^2 }$

$\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

$r = \sqrt{25.78}$

$r = 5.077 \ m$

7 0

2 years ago

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound

NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=\frac{nv_s}{4L}$

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

8 0

2 years ago

The distance of the earth from the sun is 93 000 000 miles. if there are 3.15 × 107 s in one year, find the speed of the earth i

faltersainse [42]

The angular velocity of the orbit about the sun is:

w = 1 rev / year = 1 rev / 3.15 × 10^7 s

Now in 1 rev there is 360° or 2π rad, therefore:

w = 2π rad / 3.15 × 10^7 s

To convert in linear velocity, multiply the rad /s by the radius:

v = (2π rad / 3.15 × 10^7 s) * 93,000,000 miles

<span>v = 18.55 miles / s = 29.85 km / s</span>

5 0

2 years ago

Read 2 more answers

The velocity versus time graph of particle A is tangent to the velocity versus time graph for particle B at point O. What is the

lara [203]

As velocities are tangent, the value of both Particle A and Particle B would be same for that point O (Intersecting point)

a = v / t
Here, v = 7, t = 6
So, a = 7/6
a = 1.17
As the graph is decreasing, value of acceleration would be negative.
So, a = -1.17 m/s²

In short, Your Answer would be Option C

Hope this helps!

7 0

2 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

6 0

2 years ago