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Black_prince [1.1K]
2 years ago
13

The operator of a space station observes a space vehicle approaching at a constant speed v. The operator sends a light signal at

speed c toward the space vehicle. What is the speed of the light signal as viewed from the space vehicle
Physics
1 answer:
GenaCL600 [577]2 years ago
5 0

Answer:

The speed of the light signal as viewed from the observer is c.

Explanation:

Recall the basic postulate of the theory of relativity that the speed of light is the same in ALL inertial frames. Based on this, the speed of light is independent of the motion of the observer.

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A 5-kg concrete block is lowered with a downward acceleration of 2.8 m/s2 by means of a rope. The force of the block on the Eart
maksim [4K]

When the body touches the ground two types of Forces will be generated. The Force product of the weight and the Normal Force. This is basically explained in Newton's third law in which we have that for every action there must also be a reaction. If the Force of the weight is pointing towards the earth, the reaction Force of the block will be opposite, that is, upwards and will be equivalent to its weight:

F = mg

Where,

m = mass

g = Gravitational acceleration

F = 5*9.8

F = 49N

Therefore the correct answer is E.

5 0
2 years ago
If the water vapor content of air remains constant, lowering air temperature causes _____.
Molodets [167]

<em>ANSWER</em>

<u>An increase in relative humidity</u>

<em><u>Could you mark me brainliest plz?</u></em>

8 0
2 years ago
Read 2 more answers
You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af
djyliett [7]
I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
A: mgh_1+\frac{mv_1^2}{2}
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
W_f=F_f\cdot L
So, we know that the car is approaching the point B with the following amount of energy:
mgh_1+\frac{mv_1^2}{2}- F_fL
The law of conservation of energy tells us that this energy must the same as the energy at point B. 
The energy at point B is the sum of car's kinetic and potential energy:
B: mgh_2+\frac{mv_2}{2}
As said before this energy must be the same as the energy of a car approaching the loop:
mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL
Now we solve the equation for v_1:
v_1^2=2g(h_2-h_1)+v_2^2+\frac{2F_fL}{m}\\&#10;v_1^2=39.23\\&#10;v_1=\sqrt{39.23}=6.26\frac{m}{s}

4 0
2 years ago
Read 2 more answers
Let’s now multiply two numbers in scientific notation using Google.
olga_2 [115]
We are going to rewrite each number:
 (4.48E-8) = 0.0000000448
 (5.2E-4) = 0.00052
 We observe that when multiplying, the exponent will be on the order of 10 ^ -11
 Doing the multiplication we have:
 (4.48E-8) * (5.2E-4) = 2.3296E-11
 Rewriting:
 (4.48E-8) * (5.2E-4) = 2.33E-11
 Answer:
 
2.33E-11
5 0
2 years ago
Read 2 more answers
A negative oil droplet is held motionless in a millikan oil drop experiment. What happens if the switch is opened?
scoray [572]

In Millikan oil drop experiment, when the switch is opened and by altering supply the charge of electron is determined.

Explanation:

Millikan's oil drop experiment is held to determine the terminal velocity and charge of the oil drop.

Firstly without any supply of voltage when an oil drop is sprinkled and these droplets gather electrons together and gives negative charge as they pass through air.

By applying and altering voltage applied on the plates, drop can be suspended in air. Millikan observed one drop after another, varying the voltage and noting the effect. After many repetitions he concluded that charge could assume only certain fixed values.

After conducting many times he concluded 1.602176487 ×10−19 C as the charge of an electron.

0 0
2 years ago
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