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Stolb23 [73]
3 years ago
8

A 0.065-kg tennis ball moving to the right with a speed of 15 m/s is struck by a tennis racket, causing it to move to the left w

ith a speed of 15 m/s. if the ball remains in contact with the racquet for 0.020 s, what is the magnitude of the average force exerted on the ball?
Physics
1 answer:
djverab [1.8K]3 years ago
7 0
Using the formula for impulse, we can solve this problem.
In terms of change in momentum, impulse is:
I = m (v2 - v1)

Or in terms of force and contact time:
I = F t

Equating the two equations:
m (v2 - v1) = F t

Setting the right direction as positive and the left as negative, we can then substitute values with their respective signs:
0.065 (-15 -15) = F (0.020)
F = -97.5 N

The negative sign indicates that the force was directed to the left.
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As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constant
NNADVOKAT [17]

Hi there!

a.

We can use the initial conditions to solve for w₀.

It is given that:

\frac{d\theta}{dt} = w_0e^{-\sigma t}

We are given that at t = 0, ω =  3.7 rad/sec. We can plug this into the equation:

\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}

Now, we can solve for sigma using the other given condition:

2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}

b.

The angular acceleration is the DERIVATIVE of the angular velocity function, so:

\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}

c.

The angular displacement is the INTEGRAL of the angular velocity function.

\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\

\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.

\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}

\theta = 8.471 rad

Convert this to rev:

8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}

d.

We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.

0 = 3.7e^{-0.0714t}\\\\t = \infty

Evaluate the improper integral:

\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\

\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.

\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad

Convert to rev:

51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}

8 0
3 years ago
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 3.50
kkurt [141]

Answer:

a) speed= 0.86m/s

b) Amplitude = 0.35m

C) Amplitude = 0.25m

Explanation: correct values in question (for the boat to travel from its highest point to its lowest, a total distance of 0.700 m . The fisherman sees that the wave crests are spaced 6.00 m apart.)

The speed of a periodic wave with wavelength and frequency is given by:

Speed = frequency × wavelength

The relation between the time period and frequency is given by:

T = 1/f

Given:

Time that the boat takes to travel from the highest point to its lowest point,t = 3.50s

Distance between the lowest point and the highest point,d = 0.70m

Wavelength = 6.0m

a) Wavelength is the distance between two successive wave crests or troughs. So, the distance between the Crest (highest point) to the next(lowest point) is a half wavelength.

The time period ,T is the time between two successive waves.

Therefore,the time it takes the boat to travel from its highest point to its lowest point is period.

T = 2 ×3.50 =7.0seconds

Frequency, f = 1/7.0

f = 0.1429Hz

Speed = frequency × wavelength

Speed = 0.1429 × 6 = 0.86m/s

b) The amplitude is the maximum magnitude of displacement from equilibrium.

Therefore, the distance between the boar's highest point to its lowest point is

Amplitude, A = distance /2

Amplitude A = 0.70/2= 0.35m

c) If the vertical distance changes to d2= 0.50m but other data remained the same, this wont change the speed of the wave because it does not depend on the vertical displacement but it will change the amplitude

A = distance /2 = 0.50m/2 = 0.25m

7 0
3 years ago
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In a single replacement reaction, what will a metal always replace?
Alexandra [31]
A metal have a nice day
8 0
3 years ago
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If you had a known sample of a mineral whose density was 7.1 grams/ml and its volume was 15 ml. What is the mass of the mineral.
kow [346]

Answer:

<h3>The answer is 106.5 g</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume = 15 mL

density = 7.1 g/mL

We have

mass = 7.1 × 15

We have the final answer as

<h3>106.5 g</h3>

Hope this helps you

6 0
3 years ago
A +2.00nc point charge is at the origin, and a second -5.00nc point charge is on the x-axis at x = 0.800m find the magnitude of
Damm [24]
You have to reduce 2.00 an5.00 I order to use the×that=0.800
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