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3 years ago

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Suppose you are in an elevator. As the elevator starts upward, its speed will increase. During this time when the elevator is mo

ving upward with increasing speed, your weight will be:_______.a) greater than your normal weight at rest.b) equal to your normal weight at rest.c) less than your normal weight at rest.

1 answer:

vova2212 [387]3 years ago

5 0

Answer: The answer is A.

Explanation:

Suppose you are in an elevator. As the elevator starts upward, its speed will increase. During this time when the elevator is moving upward with increasing speed, your weight will be greater than your normal weight at rest.

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A solid circular plate has a mass of 0.25 kg and a radius of 0.30 m. It starts rolling from rest at the top of a hill 10 m long

KATRIN_1 [288]

To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.

The net height from the point where it begins to roll with an inclination of 30 degrees would be

$h=Lsin30$

$h=10sin30$

$h=5m$

In the case of Inertia would be given by

$I = \frac{mR^2}{2}$

In general, given an object of mass m, an effective radius k can be defined for an axis through its center of mass, with such a value that its moment of inertia is

$I = mk^2$

$\frac{mR^2}{2}= mk^2$

$\frac{k^2}{R^2}=\frac{1}{2}$

Replacing in Energy conservation Equation we have that

Potential Energy = Kinetic Energy of Rolling Object

$mgh = \frac{1}{2}mv^2(1+\frac{k^2}{r^2})$

$9.8*5=\frac{1}{2}v^2(1+\frac{1}{2})$

$v^2 (1.5) = 98$

$v=8.0829m/s$

Therefore the correct answer is C.

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3 years ago

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

Irina-Kira [14]

Answer:

part (a) $\alpha\ =\ -2.5\ rad/s^2$

part (b) N = 79.61 rev

part (c) $\tau\ =\ 23.54\ Nm$

Explanation:

Given,

Initial speed of the wheel = $w_o\ =\ 50.0\ rad/s$
total time taken = t = 20.0 sec

part (a)

Let $\alpha$ be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

$\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2$

part (b)

Let $\theta$ be the total angular displacement of the wheel from initial position till the rest.

$\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad$

We know, 1 revolution = $2\pi$ rad

Let N be the number of revolution covered by the wheel.

$\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev$

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

Mass of the pole = m = 4 kg
Length of the pole = L = 2.5 m
Angle of the pole with the horizontal axis = $\theta\ =\ 60^o$

Now the center of mass of the pole = $d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m$

Weight component of the pole perpendicular to the center of mass = $F\ =\ mgcos\theta$

$\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm$

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4 years ago