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2 years ago

11

A diver shines an underwater searchlight at the surface of a pond (n = 1.33). At what angle (relative to the surface) will the l

ight be totally reflected?

1 answer:

allsm [11]2 years ago

4 0

Answer:

41.2°

Explanation:

Total internal reflection is the reflection of the incident ray at the interface between two media in which one of the media has a lower refractive index than the other. It occurs when the angle of incidence in the denser medium exceeds the critical angle.

The critical angle is the angle of incidence in the denser medium when the angle of incidence in the less dense medium is 90°.

Since

n= 1/sin C

C= sin^-(1/n)

C= sin^-(1/1.33)

C= 48.8°

Hence angle of incidence= 90-48.8 = 41.2°

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A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

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2 years ago

A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s

topjm [15]

Answer:

$\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})$

Explanation:

The electrostatic potential energy is given by the following formula

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Now, we will apply this formula to both cases:

$U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}$

So, the change in the potential energy is

$\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})$

7 0

2 years ago

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a

Pie

Answer:

Part a)

$v = 7407.1 m/s$

Part b)

$v_{rel} = 1.05 \times 10^4 m/s$

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

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here we know that

$M = 5.98 \times 10^{24} kg$

$R = 6.37 \times 10^6 m$

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now we have

$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}$

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Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

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2 years ago

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bearhunter [10]

Answer:

a)

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Weight in Water = 0.25N

Weight in Liquid = 0.24N.

Upthrust /Buoyant Force = Weight in Air – Weight in Fluid(Water in this case)

= 0.3 – 0.25

= 0.5N.

b) R.D of Body = Density of Body/Density of Standard Fluid(Water).

There's a Derived Formula for RD.

I'm gonna Apply it here.

Ask me for the derivation in the Comment section if you need it.

RD = α/ρ = (Weight in Air) / (Upthrust Force)

Where

α = density of the Body(or reference substance)

ρ = density of standard fluid (water)

= 0.3/0.05 = 6.

c) RD of Liquid = (Density of Liquid) /(Density of standard Fluid(water)

Or we just go by that formula

RD of Liquid = Weight in Air/Upthrust(In Liquid)

We'll be using the Upthrust in that Liquid now.

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7 0

1 year ago

M A sinusoidal wave on a string is described by the wave function

IceJOKER [234]

The frequency of the wave is determined as 7.96 Hz.

<h3>Frequency of the wave</h3>

The frequency of the wave is calculated as follows;

y = A sin(ωt - kx)

where;

A is amplitude of the wave
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ω = 2πf

f = ω/2π

f = (50)/(2π)

f = 7.96 Hz

Thus, the frequency of the wave is determined as 7.96 Hz.

Learn more about frequency of waves here: brainly.com/question/6297363

#SPJ4

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1 year ago