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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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What's the difference between Uniform magnetic field and Non-uniform magnetic field?
Galina-37 [17]
Magnetic field is the magnetic effect of electric currents and magnetic materials. The magnetic field at any given point specified by both direction and a magnitude. So uniform magnetic field has equal amount of force or magnetic force in both side of the object while the non uniform magnetic field has one of the object exert more than the other 
4 0
3 years ago
A steel cable has a cross-sectional area 4.49 × 10^-3 m^2 and is kept under a tension of 2.96 × 10^4 N. The density of steel is
Lemur [1.5K]

Answer:

The transverse wave will travel with a speed of 25.5 m/s along the cable.

Explanation:

let T = 2.96×10^4 N be the tension in in the steel cable, ρ  = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.

then, if V is the volume of the cable:

ρ = m/V

m = ρ×V

but V = A×L , where L is the length of the cable.

m = ρ×(A×L)

m/L = ρ×A

then the speed of the wave in the cable is given by:

v = √(T×L/m)

  = √(T/A×ρ)

  = √[2.96×10^4/(4.49×10^-3×7860)]

  = 25.5 m/s

Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.

7 0
3 years ago
A pure musical torte causes a thin wooden
victus00 [196]
Hello! Your answer would be D. Interference

This is because interference causes vibration!!!
4 0
3 years ago
Two 60 cm parallel disks are separated by 40 cm and are aligned directly on top of each other. Both disks are black surfaces wit
Crazy boy [7]

Answer:

775.48 W

Explanation:

given,

diameter of disk = 0.6 cm

length of the disk = 0.4 m

T₁ = 450 K         T₂ = 450 K      T₃ = 300 K

\dfrac{d}{r_1}=\dfrac{0.4}{0.3} = 1.33

now,

the value of view factor (F₁₂)corresponding to 1.33

F₁₂ = 0.265

F₁₃ = 1 - 0.265 = 0.735

now,

net rate of radiation heat transfer from the disk to the environment:

=\dot{Q_{1-3}+Q_{2-3}} = 2 \dot{Q_{1-3}}

       = 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)

       = 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)

       = 775.48 W

Net radiation heat transfer from the disks to the environment = 775.48 W

3 0
4 years ago
The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20
IrinaK [193]

Answer:

The average angular acceleration of the Earth is; α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

We are given;

The period of 365 revolutions of Earth in 2006, T₁ = 365 days, 0.840 sec

Converting to seconds, we have;

T₁ = (365 × 24 × 60 × 60) + 0.84

T₁ = (3.1536 x 10⁷) + 0.840

T₁ = 31536000.84 s

Now, the period of 365 rotation of Earth in 2006 is; T₀ = 365 days

Converting to seconds, we have;

T₀ = 31536000 s

Hence, time period of one rotation in the year 2006 is;

Tₐ = 31536000.84/365

Tₐ = 86400.0023 s

The time period of rotation is given by the formula;

Tₐ = 2π/ωₐ

Making ωₐ the subject;

ωₐ = 2π/Tₐ

Plugging in the relevant values;

ωₐ = 2π/ 365.046306        

ωₐ = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation in the year 1906 is;

Tₓ = 31536000/365

Tₓ = 86400 s

Time period of rotation,

Tₓ = 2π /ωₓ

ωₓ = 2π / T

Plugging in the relevant values;

ωₓ = 2π/86400

ωₓ = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration is given by;

α  = (ωₓ -   ωₐ) /  T₁

α = ((7.272205217  × 10⁻⁵) - (7.272205023 × 10⁻⁵)) / 31536000.84

 α  = 6.152 X 10⁻²⁰ rad/s²

Thus, the average angular acceleration of the Earth is; α  = 6.152 X 10⁻²⁰ rad/s²

8 0
4 years ago
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