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bogdanovich [222]
3 years ago
13

Which postulate of relativity was supported by the experiments of Michelson and Morley

Physics
2 answers:
shtirl [24]3 years ago
6 0
Constancy of speed of light.
Flauer [41]3 years ago
5 0
All observers measure the same speed of light,
no matter how any observer is moving.

The question is kind of awkward, though.  It seems to say that
people were having a hard time accepting this postulate of relativity,
until Michelson and Morley came along and did their experiment, and
their results gave Relativity a boost.

Actually, Michelson and Morley did their famous experiment in 1887.
At the time, Einstein was only 8 years old, and Relativity was not even
a twinkle in his eye yet.
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When a swimmer pushes through water to swim they are propelled forward because of the water resistance against the hand and feet. It's A. The water doesn't automatically push the swimmer forward. It releases a reaction after the swimmer pushes through the water.
6 0
3 years ago
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a hockey puck with a mass of 0.11 kg is at rest on the horizontal frictionless surface of the rink. a player applies a horizonta
stira [4]

The solution to this ques is available in the image.

Given,

Force= 1N

Mass= 0.11kg

Time= 5sec

Force= mass X accelaration

Accelaration= velocity/ time

Speed=distance/ time

Hence, the speed is 45 m/s and the distance is 225 m.

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8 0
1 year ago
A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m
Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

a=2.92164g

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s

Angular velocity is 1.3823 rad/s

Linear velocity is given by

v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g

a=2.92164g

Centripetal force is given by

F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0
3 years ago
Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 s to speed up from rest t
Vinvika [58]

Answer

Time period T = 1.50 s

time t = 40 s

r = 6.2 m

a)

Angular speed ω = 2π/T

                              = \dfrac{2\pi }{1.5}  

                              = 4.189 rad/s

Angular acceleration α = \dfrac{\omega}{t}

                                      = \dfrac{4.189}{40}

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Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²

b)The maximum speed.

       v = 2πr/T

          = \dfrac{2\pi \times 6.2}{1.5}

          = 25.97 m/s

So centripetal acceleration.

        a = \dfrac{v^2}{r}

          = \dfrac{25.97^2}{6.2}

          =  108.781 m/s^2

          = 11.1 g    

in combination with the gravitation acceleration.

a_{total} = \sqrt{(11.1g)^2+g^2}

a_{total}= 11.145 g

6 0
3 years ago
Joanne drives her car at a speed of 20 m/s. when she applied her breaks, a frictional force of 2000 N brought her car to a compl
Papessa [141]

Answer:

A) 1000 kg

Explanation:

vf = vi + at

0 = 20 + (a)(10)

a = -2.0 m/s^2

F = ma

2000 = (m)(2)

m = 1000 kg

8 0
1 year ago
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