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3 years ago

According to Einstein's special theory of relativity, in which frames of reference will all laws of physics apply?

Physics

2 answers:

soldier1979 [14.2K]3 years ago

5 0

Answer:

A. Within any frame of reference that is not accelerating.

Explanation:

Just answered it on Apex!

kenny6666 [7]3 years ago

3 0

Answer: Within any frame of reference that is accelerating

Special relativity was proposed on 1905 by Einstein, who developed his theory based on the following two postulates:

1. The laws of physics are the same in all inertial systems. There is no preferential system.

2. The speed of light in vacuum has the same value for all inertial systems.

Focusing on the first postulate, it can be affirmed that any measurement on a body is made with reference to the system in which it is being measured.

Now, taking into account that an inertial reference system is the one that complies with the principle of inertia:

"For a body to have acceleration, an external force must act on it."

The correct answer is

Within any frame of reference that is accelerating

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A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is

valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

$\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2$

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

$\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}$

= 4.7476 m/sec

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6 0

2 years ago

There is a naturally occurring vertical electric field near the Earth’s surface that points toward the ground. In fair weather c

trasher [3.6K]

Answer:

$\frac{F}{W} = 9.37 \times 10^{-4}$

Explanation:

Radius of the pollen is given as

$r = 12.0 \mu m$

Volume of the pollen is given as

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi (12\mu m)^3$

$V = 7.24 \times 10^{-15} m^3$

mass of the pollen is given as

$m = \rho V$

$m = 7.24 \times 10^{-12}$

so weight of the pollen is given as

$W = mg$

$W = (7.24 \times 10^{-12})(9.81)$

$W = 7.1 \times 10^{-11}$

Now electric force on the pollen is given

$F = qE$

$F = (-0.700\times 10^{-15})(95)$

$F = 6.65 \times 10^{-14} N$

now ratio of electric force and weight is given as

$\frac{F}{W} = \frac{6.65 \times 10^{-14}}{7.1 \times 10^{-11}}$

$\frac{F}{W} = 9.37 \times 10^{-4}$

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2 years ago

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Answer:

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