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4 years ago

13

What is the wavelength of the light entering an interferometer if 402 bright fringes are counted when the movable mirror moves 0

.115 mm?

1 answer:

rusak2 [61]4 years ago

8 0

Answer:

The value is $\lambda = 572 *10^{-9} \ m$

Explanation:

From the question we are told that

The number of bright fringes is n = 402

The displacement of the mirror is $\delta l = 0.115 \ mm = 1.15 *10^{-4} \ m$

Generally the number of fringes is mathematically represented as

$n = \frac{2 * \delta l }{\lambda }$

=> $\lambda = \frac{2 * \delta l }{n }$

=> $\lambda = \frac{2 * 1.15*10^{-4} }{402 }$

=> $\lambda = 572 *10^{-9} \ m$

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A .100 kg bullet is flying at 200 m/s. How much energy is the bullet storing due to its motion?

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KE = 2000 J

Explanation:

KE = (1/2)mv^2

= (1/2)(0.100 kg)(200 m/s)^2

= 2000 J

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3 years ago

An astronaut holds a rock 100m above the surface of Planet X. The rock is then thrown upward with a speed of 15m/s, as shown in

Harlamova29_29 [7]

The acceleration due to gravity of the planet X is 1 m/s².

The given parameters;

height above the ground, h = 100 m
initial velocity of the rock, u = 15 m/s
time of motion of the rock, t = 10 s

The acceleration due to gravity is calculated as follows;

$h = ut - \frac{1}{2} gt^2\\\\100 = 15(10) - (0.5\times 10^2)g\\\\100 = 150 - 50g\\\\50g = 150-100\\\\50g = 50\\\\g = 1 \ m/s^2$

Thus, the acceleration due to gravity of the planet X is 1 m/s²

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2 years ago

A middle-A tuning fork vibrates with a frequency f of 440 hertz (cycles per second). You strike a middle-A tuning fork with a fo

jeyben [28]

Answer:

P = 5sin(880πt)

Explanation:

We write the pressure in the form P = Asin2πft where A = amplitude of pressure, f = frequency of vibration and t = time.

Now, striking the middle-A tuning fork with a force that produces a maximum pressure of 5 pascals implies A = 5 Pa.

Also, the frequency of vibration is 440 hertz. So, f = 440Hz

Thus, P = Asin2πft

P = 5sin2π(440)t

P = 5sin(880πt)

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3 years ago

Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

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3 years ago

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Answer:

The Doppler Effect is given by the following relation;

$f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f$

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

$v_o$ = The velocity of the observer

$v_s$ = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(ii) When the source is receding from the observer, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

(1) Given that (v + $v_s$ ) > v, therefore, v < (v + $v_s$ ), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(1) Given that (v - $v_s$ ) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

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3 years ago