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Anna71 [15]
3 years ago
11

We know theory of relativity and refraction too.Theory of relativity states that speed of light is constant but according to ref

raction, the speed of light changed when it change its medium.Then which one is true if both then how???Plz answer fast
Physics
1 answer:
Ksivusya [100]3 years ago
6 0
Theory of relativity shows that the speed of light doesn't depend
on the motion of the observer.  All observers measure the same
speed of light no matter how they're moving.

It does NOT say that the speed of light is the same in any medium.
If the light is moving through a block of jello, then all observers in jello
will measure the same speed of light, no matter how they're moving
and whatever that speed is.
You might be interested in
What is the height of a building that an object is dropped from if it has a mass of 3 kg and hits the ground with a velocity of
Marizza181 [45]

Answer:

125 m

Explanation:

m = 3kg

v = 50m/s

u = 0m/s

a = +g = 10m/s²

s = H = ?

using the formula,

v² = u² + 2as

50² = 0² + 2(10)(H)

2500 = 20H

H = 2500/20

H = 125m

8 0
3 years ago
A wheel of diameter 78 cm has an axle of diameter 14.8 cm. A force 150 N is exerted along the rim of the wheel.What force should
bezimeni [28]

Answer:

F = 263.51 N

Explanation:

given,

diameter of wheel = 78 cm

diameter of axle = 14.8 cm

Force exerted on the rim of wheel = 150 N

Force applied outside the axle = ?

To prevent rotation wheel from rotating the Force 'F' should be applied outside of the axle.

Net momentum about the center of mass should be zero

now,

Moment of about center due to 150 N = moment about center due to F on axle

50\times \dfrac{78}{2}=F\times \dfrac{14.8}{2}

           7.4 F = 1950

                F = 263.51 N

Hence, Force exerted outside of the axle in order to prevent the wheel from rotating is equal to 263.51 N.

6 0
4 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
4 years ago
The cylinder of gravity of cylinder is where
Elodia [21]

Explanation:

In uniform gravity it is the same as the centre of mass. For regular shaped bodies it lies at the centre of the that particular body. Hence for a cylinder centre of gravity lies at the midpoint of the axis of the cylinder.

5 0
4 years ago
Read 2 more answers
A turbine blade rotates with angular velocity ω(t) = 8.00 rad/s - 0.20 rad/s3 t 2. what is the angular acceleration of the blade
Irina18 [472]
Given: <span>ω(t) = 8.00 - 0.2*t^2

Differentiating both sides with respect to time we get:
angular acceleration = </span>α(t) = -0.2 * 2 *t 

At t = 6.8 sec,
angular acceleration = α(t) = -0.2 * 2 *6.8 = -2.72 rad/s^2
4 0
4 years ago
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