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3 years ago

Differences between Constant velocity and constant acceleration

Physics

1 answer:

BartSMP [9]3 years ago

3 0

Answer:

Traveling with a constant velocity means you're going at the same speed in the same direction continuously. If you have a constant velocity, this means you have zero acceleration. ... If you travel with a constant acceleration, your velocity is always changing, but it's changing by a consistent amount each second.

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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!

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:

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

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

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

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:

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

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 race car completed a race in 150 minutes. Its average speed was found to be 150 km/hr. What is the total distance that the car

hichkok12 [17]

375 kilometers because 150 minutes is 2.5 hours

3 0

3 years ago

Read 2 more answers

How much force does it take to accelerate a 3.50 kg mass from 3.00 m/s to 17.0 m/s in a time of 2.00 seconds?

zheka24 [161]

Answer:

Force = 24.5 Newton

Explanation:

Given the following data;

Mass = 3.50 kg

Initial velocity, u = 3 m/s

Final velocity, v = 17 m/s

Time, t = 2 seconds

To find the force;

First of all, we would determine the acceleration of the object using the formula;

Acceleration = (v - u)/t

Acceleration = (17 - 3)/2

Acceleration = 14/2

Acceleration = 7 m/s²

Now, we can find the force using the formula;

Force = mass * acceleration

Force = 3.5 * 7

Force = 24.5 Newton

5 0

4 years ago

Puck 1 (1 kg) travels with velocity 30 m/s to the right when it collides with puck 2 (1 kg) which is initially at rest. After th

Katyanochek1 [597]

Answer:

v₂ = 22.5 m/s

Explanation:

Given that

For puck 1

m₁= 1 kg

u₁= 30 m/s

For puck 2

m₂= 1 kg

u₂= 0 m/s

After collision

Puck 1 have velocity v₁=7.5 m/s

Take puck 2 will have velocity v₂

From linear momentum conservation

P₁=P₂

m₁ u₁+m₂ u₂=m₁ v₁+m₂ v₂

1 x 30 + 1 x 0 = 1 x 7.5 + 1 x v₂

30 - 7.5 =v₂

v₂ = 22.5 m/s

6 0

3 years ago

You have a stopped pipe of adjustable length close to a taut 62.0 cmcm, 7.25 gg wire under a tension of 4710 NN. You want to adj

Oksana_A [137]

Answer:

6 cm long

Explanation:

F = 4110N

Vo(speed of sound) = 344m/s

Mass = 7.25g = 0.00725kg

L = 62.0cm = 0.62m

Speed of a wave in string is

V = √(F / μ)

V = speed of the wave

F = force of tension acting on the string

μ = mass per unit density

F(n) = n (v / 2L)

L = string length

μ = mass / length

μ = 0.00725 / 0.62

μ = 0.0116 ≅ 0.0117kg/m

V = √(F / μ)

V = √(4110 / 0.0117)

v = 592.69m/s

Second overtone n = 3 since it's the third harmonic

F(n) = n * (v / 2L)

F₃ = 3 * [592.69 / (2 * 0.62)

F₃ = 1778.07 / 1.24 = 1433.927Hz

The frequency for standing wave in a stopped pipe

f = n (v / 4L)

Since it's the first fundamental, n = 1

1433.93 = 344 / 4L

4L = 344 / 1433.93

4L = 0.2399

L = 0.0599

L = 0.06cm

L = 6cm

The pipe should be 6 cm long

6 0

3 years ago