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3 years ago

The vertical displacement of the wave is measured from the ?

1 answer:

6 0

The whole question is talking about the amplitude of a wave
that's transverse and wiggling vertically.

Equilibrium to the crest . . . that's the amplitude.

Crest to trough . . . that's double the amplitude.

Trough to trough . . . How did that get in here ? Yes, that's
   the wavelength, but it has nothing to do
   with vertical displacement.

Frequency . . . that's how many complete waves pass a mark
   on the ground every second. Doesn't belong here.

Notice that this has to be a transverse wave. If it's a longitudinal wave,
like sound or a slinky, then it may not have any displacement at all
across the direction it's moving.

It also has to be a vertically 'polarized' wave. If it's wiggling across
the direction it's traveling BUT it's wiggling side-to-side, then it has
no vertical displacement. It still has an amplitude, but the amplitude
is all horizontal.

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A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

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Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

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Answer:F=4F

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Force between the two charges is given by

F=K*q1*q2/r^2

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F=4(K*q1*q2)/r^2

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