Please help me with these two multiple choice questions!

A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a ra

worty [1.4K]

Answer:

The torque on the pulley, when the system is motionless is approximately 9.81 N·m

Explanation:

The given parameters are;

The mass of the object = 2 kg

The friction between the rope and the pulley = 0

The mass of the rope, $m_r$ = 0.5 kg

The mass of the pulley, $m_p$ = 0.01 kg

The radius of the pulley, r = 0.25 m

The torque on the pulley, τ = I·α = F × D

The torque on the pulley, when the system is motionless, τ = F × D

Where;

F = The force acting on the pulley rope = The weight of the mass ≈ 2 kg × 9.81 m/s² = 19.62 N

D = The diameter of the pulley = 2×r = 2 × 0.25 m = 0.5 m

Therefore;

τ = 19.62 N × 0.5 m = 9.81 N·m

The torque on the pulley, when the system is motionless, τ ≈ 9.81 N·m.

4 0

3 years ago

Select all the wave properties that can change how we perceive light waves.amplitudeamplitudecrestscreststimetimefrequency

ASHA 777 [7]

The properties that change how we perceive light waves are the following:

The amplitude of the light wave changes the brightness of light relative to other light waves of the same wavelenghth.

The frequency of the light wave changes the color and the type of the light wave.

6 0

1 year ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

$\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m$

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

$y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)$

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

$v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)$

$a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0$

For this equation to be equal to zero, sin(59.94t) = 0. So,

$59.94t = \pi\\t = \pi/59.94 = 0.0524~s$

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

$v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s$

4 0

3 years ago

Imagine that while you and a passenger are in a deep-diving submersible in the North Pacific near Alaska’s Aleutian Islands, you

mamaluj [8]

Answer:

I would say its a deep ocean trench

Explanation:

This is because deep ocean trenches are found at the deepest part of the ocean and also at Pacific ocean margins or Rim where subduction usually occurs and Aleutian islands are part of the Pacific Rim

6 0

3 years ago

The baseball player is sliding towards the left of the picture to get to second base. In what direction is the force of friction

lubasha [3.4K]

The answer is B. Friction is going to the RIGHT because friction works against where you are trying to go

6 0

3 years ago

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