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Wewaii [24]
3 years ago
10

A rocket is moving at 1/4 the speed of light relative to Earth. At the center of this rocket a light suddenly flashes. To an obs

erver at rest in the rocket a. the light will reach the front of the rocket before it reaches the back of the rocket. b. the light will reach the front of the rocket at the same instant that it reaches the back of the rocket. c. the light will reach the front of the rocket after it reaches the back of the rocket.
Physics
2 answers:
larisa [96]3 years ago
7 0

Answer:

The correct answer is;

b. the light will reach the front of the rocket at the same instant that it reaches the back of the rocket.

Explanation:

The speed of light is an invariant. The speed of light is independent of the source or the observer, that is the speed of light does not depend on the speed of the source, or the speed of the observer, or the speed of the medium. The speed of light is constant, and as such the light will arrive at the back and front of the rocket at the same time.

Sedaia [141]3 years ago
5 0

Answer:

B. the light will reach the front of the rocket at the same instant that it reaches the back of the rocket.

Explanation:

To an observer at rest in the rocket who can't see either sides of the rocket, the speed of the light is constant which means the distance to the front or the back is same and would appear to reach the rocket at the same time.

Although from the point of view of the person on the earth, the front of the rocket is travelling in opposite direction of the light while the back of the rocket is moving closer to the light. This means that the distance travelled by the light going forward will be longer going backwards. And since the speed of light is constant in both directions, the light will reach the back of the rocket before it reaches the front  for the observer on the earth.

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A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What
kati45 [8]

Explanation:

The given data is as follows.

           radius (r) = 3.25 cm,    \alpha = 11.6 rad/s^{2}

Now, we will calculate the tangential acceleration as follows.

          a_{tangential} = \alpha \times r

Putting the given values into the above formula as follows.

         a_{tangential} = \alpha \times r

                      = 11.6 rad/s^{2} \times 3.25 cm

                      = 37.7 rad cm/s^{2}

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 rad cm/s^{2}.

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3 years ago
How much time would it take for the sound of thunder to travel 2000 meters if sound travels of 330 meters per sec
Lubov Fominskaja [6]
2000÷330=6.06 repatant so the answer would be about 6.06 seconds
4 0
3 years ago
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QUESTION 36
mel-nik [20]

Answer:

Explanation:

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An l-c cirucit with a 70 mh inductor and a ,54 F capacitor oscillates. The maximum charge on the capacitor is 11.5 C. What are t
Oxana [17]

This question is incomplete, the complete question is;

An l-c cirucit with a 70 mh inductor and a 0.54 μF capacitor oscillates. The maximum charge on the capacitor is 11.5 μC. What are the oscillation frequency and the maximum current in this circuit ;

Options

a) 1.07 kHz, 63.4 mA

b) 4.38 kHz, 101.3 mA

c) 6.74 kHz, 55.7 mA

d) 2.31 kHz, 93.5 mA

e) 0.82 kHz, 59.1 mA

Answer:

the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

Explanation:

Given that;

inductor L = 70 mH = 70 × 10⁻³ H

Capacitor C = 0.54 μf = 0.54 × 10⁻⁶ f

Qmax on capacitor = 11.5 μf = 11.5 × 10⁻⁶ c

oscillation frequency in L-C circuit;

f = 1/2π√(LC)

we substitute our values;

f = 1/2π√(70 × 10⁻³ × 0.54 × 10⁻⁶  )

f = 0.0818 × 10⁴ Hz

f = 0.082 × 10³ Hz ≈ 0.82 kHz

Maximum circuit in L-C circuit is given by

I_max = Qmax/√(LC)

we substitute

I_max = 11.5 × 10⁻⁶ / √(70 × 10⁻³ × 0.54 × 10⁻⁶  )

= 59.1 × 10³ A ≈  59.1 mA

Therefore the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

5 0
2 years ago
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