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katovenus [111]
3 years ago
11

The Michelson-Morley experiment a) confirmed that time dilation occurs. b) proved that length contraction occurs. c) verified th

e conservation of momentum in inertial reference frames. d) supported the relationship between mass and energy. e) indicated that the speed of light is the same in all inertial reference frames.
Physics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

e) indicated that the speed of light is the same in all inertial reference frames.

Explanation:

In 18th century, many scientists believed that the light just like air and water needs a medium to travel. They called this medium <em>aether</em>. They believed that even the space is not empty and filled with aether.

Michelson and Morley tried to prove the presence and speed of this aether through an interference experiment in 1887. They made an interferometer in which light was emitted at various angles with respect to the supposed aether. Both along the flow and against the flow to see the difference in the speed of light. But they did not find no major difference and thus it became the first proof to disprove the theory of aether.

It thus proved that the speed of light remains same in all inertial frames.

Also, it became a base for the special theory of relativity by Einstein.

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Which of the following statements are true?
inessss [21]

Answer:

a. If an object's speed is constant, then its acceleration must be zero.

FALSE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

so we can not say anything about the acceleration when speed is given to as and no information is given about velocity

b. If an object's acceleration is zero, then its speed must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

Since we know that if acceleration is 0 then velocity must be constant and hence speed is also constant

c. If an object's velocity is constant, then its speed must be constant.

TRUE

Since velocity is constant then it shows that its magnitude and direction both are constant so its speed is also constant.

d. If an object's acceleration is zero, its velocity must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

Since we know that if acceleration is 0 then velocity must be constant

e. If an object's speed is constant, then its velocity must be constant.

FALSE

Speed is just the magnitude so we can not say about its direction and hence if speed is constant then velocity may or may not change

7 0
3 years ago
Thermal energy depends on an object’s
oee [108]

The correct answer is D. I alread took this test.

4 0
3 years ago
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High energy waves have what
GarryVolchara [31]
High energy waves have Gamma rays
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3 years ago
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Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ
Tomtit [17]

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}

Put the value into the formula

525=\sqrt{\dfrac{c+v}{c-v}}\times950

\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2

\dfrac{c+v}{c-v}=0.305

c+v=0.305\times(c-v)

v(1+0.305)=c(0.305-1)

v=\dfrac{0.305-1}{1+0.305}c

v=−0.532c

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

7 0
3 years ago
Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro
aniked [119]

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

S = \frac{Q}{T}

Here,

Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy \Delta S_{hot} is smaller than \Delta S_{cold}

Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object

5 0
3 years ago
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