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3 years ago

Hold the paper up to a mirror so

Physics

2 answers:

Ne4ueva [31]3 years ago

5 0

Answer:

The image of everything in front of the mirror is reflected backward, retracing the path it traveled to get there. Nothing is switching left to right or up-down. Instead, it's being inverted front to back. ... That reflection represents the photons of light, bouncing back in the same direction from which they came

Explanation:

SpyIntel [72]3 years ago

5 0

Explanation:

there is a mirror reflection

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It would be 3.15 in meters

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3 years ago

What do Babys Come From

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Answer:

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4 years ago

Let theta denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the Bob is m, the

posledela

The tension has to hold the part of the weight in the direction of the string:

T = mg*cos(theta)

Theta=0, whole weight, theta=90, T=0, if the pendulum is horizontal, the string will be loose! Yeah

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3 years ago

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In the figure, a weightlifter's barbell consists of two identical small but dense spherical weights, each of mass 50 kg. These w

kondaur [170]

The moment of inertia is $24.8 kg m^2$

Explanation:

The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.

The moment of inertia of the rod about its centre is given by

$I_r = \frac{1}{12}ML^2$

where

M = 24 kg is the mass of the rod

L = 0.96 m is the length of the rod

Substituting,

$I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2$

The moment of inertia of one ball is given by

$I_b = mr^2$

where

m = 50 kg is the mass of the ball

$r=\frac{L}{2}=\frac{0.96}{2}=0.48 m$ is the distance of each ball from the axis of rotation

So we have

$I_b = (50)(0.48)^2=11.5 kg m^2$

Therefore, the total moment of inertia of the system is

$I=I_r + 2I_b = 1.84+ 2(11.5)=24.8 kg m^2$

Learn more about inertia:

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3 years ago

The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

3 0

3 years ago

Read 2 more answers