Dont mind this question

What is the ideal mechanical advantage of the pulley system shown in this figure?

olchik [2.2K]

Answer:

v

Explanation:

vv

4 0

3 years ago

What amount of force is required for a 200 kg grizzly bear to accelerate at 2.5 m/s/<br> s?

Len [333]

Dont mind the top part, hope this helped

5 0

3 years ago

The image represents a sound wave as it goes through several changes. Describe how the waves changes. where does it get louder?

Ksju [112]

Answer:

At point A. it gets kinda loud then at point B. it gets very loud than at point C. it gets quieter than at point D. it gets more silent.

Explanation:

Because you can tell when the waves get large it gets louder and then when the waves get more far apart then it get more quiet. Hope this helped!

4 0

3 years ago

For small angles, does the pendulum's period of oscillation depend on initial angular displacement from equilibrium? Explain.

aksik [14]

Answer:

No, the pendulum's period of oscillation does not depend on initial angular displacement.

Explanation:

Given that,

For small angle, the pendulum's period of oscillation depend on initial angular displacement from equilibrium.

We know that,

The time period of pendulum is defined as

$T=2\pi\sqrt{\dfrac{l}{g}}$

Where, l = length of pendulum

g = acceleration due to gravity

So, The time period of pendulum depends on the length of pendulum and acceleration due to gravity.

It does not depend on the initial angular displacement.

Hence, No, the pendulum's period of oscillation does not depend on initial angular displacement.

6 0

4 years ago

A 250 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.5 kg tr

Lady_Fox [76]

Answer:

$\omega_{t}=-3.9 rpm$

Explanation:

Let's use the conservation of angular momentum here. If there is a conservation the addition of both must be zero.

$L_{c}+L_{t}=0$

The momentum of inertia of both are:

$I_{c}=m_{c}R^{2}$

$I_{t}=m_{t}R^{2}$

The angular momentum is the product between angular velocity and momentum of inertia.

$I_{c}\omega_{c}+I_{t}\omega_{t}=0$

Let's solve it for ω(t).

$\omega_{t}=-\frac{I_{c}\omega_{c}}{I_{t}}$

$\omega_{t}=-\frac{m_{c}R^{2}\omega_{c}}{m_{t}R^{2}}$

But $\omega=V/R$

R is the radius R=D/2 = 30 cm = 0.3 m
V is the steady speed V = 0.74 m/s

$\omega_{t}=-\frac{m_{c}R^{2}*V/R}{m_{t}R^{2}}$

$\omega_{t}=-\frac{m_{c}V}{m_{t}R}$

$\omega_{t}=-\frac{0.25*0.74}{1.5*0.3}$

$\omega_{t}=-0.41 rad/s$

Knowing that 2π rad is a rev. We have.

$\omega_{t}=-3.9 rpm$

The minus sign means the track is moving opposite of the car.

I hope it helps you!

4 0

4 years ago

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