PLEASE HELP ASAPPPPP

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

Calculate (A⃗ ×B⃗ )⋅C⃗ for the three vectors A⃗ with magnitude A = 4.86 and angle θA = 23.5 ∘ measured in the sense from the +x

trasher [3.6K]

Answer:

(A⃗ ×B⃗ )⋅C⃗ = 69.868

Explanation:

We simplify the cross product first, thereafter the solution of the cross product is now simplified with the dot product as shown in the step by step calculation in the attachment

4 0

3 years ago

A real gas is changed slowly from state 1 to state 2. During this process no work is done on or by the gas . This process must b

umka2103 [35]

Answer:

Isovolumic

Explanation:

In thermodynamics, the process whereby A real gas is changed slowly from state 1 to state 2 and in this process there's no work done on or by the gas is called Isovolumic process.

5 0

3 years ago

What is specific latent heat mean in simple word?

Nuetrik [128]

Answer:

Amount of energy required to change the 1kg of a substance without changing its temperature.

7 0

2 years ago

What force is necessary to pull a 10 kg sled across a horizontal surface at a constant velocity if the coefficient of kinetic fr

Igoryamba

Answer:

Any force greater than 24.5 N

Explanation:

To find the frictional force is the 1st step. this can be found by multiplying the coefficient of friction and the normal force. normal force can be found by multiplying gravity (we will say 9.8 m/s/s) and the mass which is 10. You then multiply the result by the coefficient of friction which is 0.25 and it leads us to an answer to 24.5 N. This means that if you pull with exactly 24.5 N, the sled wouldnt move. So you need a force greater than 24.5 N and we have our answer.

I hope I helped correct me if im wrong

3 0

3 years ago

PLEASE HELP ASAPPPPP​

PLEASE HELP ASAPPPPP