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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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while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity o
MariettaO [177]

Complete question:

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency​ reflected off the wall to the bat?

Answer:

The frequency reflected by the stationary wall to the bat is 41 kHz

Explanation:

Given;

frequency emitted by the bat, = 39 kHz

velocity of the bat, v_b = 8.32 m/s

speed of sound in air, v = 340 m/s

The apparent frequency of sound striking the wall is calculated as;

f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz

The frequency reflected by the stationary wall to the bat is calculated as;

f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz

f_s\approx 41 \ kHz

3 0
3 years ago
Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a
____ [38]

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, v=2.79\times 10^8\ m/s

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s

Time period of oscillation measured by the observer is :

t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s

So, the time period of oscillation measured by the observer is 5.79 seconds.

3 0
3 years ago
Rahul wants to change the motion map shown so that it shows uniform circular motion. What change should Rahul make?
saw5 [17]

Answer:

B. He should change the lengths of the vectors that point tangent to the circle so that each is the same length.

Explanation:

A uniform circular motion is a motion in a circle where the tangential speed of the object is constant.

In the motion map:

- The arrows pointing towards the centre of the circle represent the centripetal acceleration, and their length represent the magnitude of the acceleration

- The arrows pointing tangential to the circle represent the tangential speed, and their length represent the magnitude of the speed

In this motion map, we see that the length of the vectors pointing tangent to the circle is not constant: this means that the speed is not constant. In order to have a uniform circular motion, the speed must be constant, therefore the lengths of the vectors that point tangent to the circle must be the same.

6 0
3 years ago
Read 2 more answers
The additional product of the nuclear fission reaction shown in the
Arada [10]
The answer to your question is C.
8 0
4 years ago
When you bent the plastic ruler, what did you observe in its size? ​
WARRIOR [948]

Answer:

The size will increase.

Explanation:

When you bend a plastic ruler, it's size will increase because it is elastic and will exhibit elastic deformation. When it is been bent, it will continue to stretch until it get to a point where it will not be able to regain it formal shape, it size wound of increase. Therefore when the ruler get to elastic limit and you have bend it to the point it cannot regain it's formal shape back, it will remain bent and if further force is apply on it,it will break.

3 0
4 years ago
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