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neonofarm [45]
4 years ago
11

Einstein’s general theory of relativity made or allowed us to make predictions about the outcome of several experiments that had

not yet been carried out at the time the theory was first published. Describe three experiments that verified the predictions of the theory after Einstein proposed it.
Physics
1 answer:
PtichkaEL [24]4 years ago
3 0

Answer:

1. bending of light in gravitational fields.

2. effect of gravitational redshift.

3. perihelion precission of mecury.

Explanation:

1 bending of light in gravitational fields, we can think of it like this:

by noting the change in position s of stars as they pass near the sun on the celetial sphere, so since the sun creates a gravitational field even the star thats not in our line of side(behind the sun) can be seen because its light is bent.

2. effects of gravitational redshift:

this says that if you are in the gravitational field, your clock moves slower when it is seen by a distant observer.

3. perihelion precission of mecury:

according to Newtonian physics a two body system consisting of a lone orbiting the spherical mass would trace out an ellipse with the center of mass of the system as the focus but mercury deviates from that precission. then according to Einstein, the change in orientation of the orbital ellipsewithin its orbital plane is the effect of gravitation being mediated by the curvature of space-time.

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Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o
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Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

U = \frac{kq_{1}q_{2}  }{r}

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

U_{1}  = \frac{ke^{2}   }{r_{1} }

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

U_{2}  = \frac{ke^{2}   }{r_{2} }=\frac{ke^{2}   }{4.10r_{1} }

Applying law of conservation of energy :

ΔU  = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁  - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

\frac{1}{2}mv^{2}=\frac{ke^{2}   }{r_{1} }- \frac{ke^{2}   }{4.10r_{1} }

Here m and v are the mass and final speed of electron respectively.

v^{2}=\frac{2}{m} \frac{ke^{2}   }{r_{1} }(1- \frac{1  }{4.10 })

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2}   }{4.32\times10^{-10} }(1- \frac{1  }{4.10 })

v^{2}=8.86\times10^{11}

v = 9.41 x 10⁵ m/s

5 0
3 years ago
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