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2 years ago

Which best describes what Einstein’s theories tried to explain?

Physics

2 answers:

salantis [7]2 years ago

6 0

Answer:

how the physical world functions

Explanation:

It's easy enough to state the bare essence of the theory in a couple pithy statements: "Matter and energy tell space-time how to bend, and the bending of space-time tells matter how to move." But the actual mechanics take a whopping 10 equations to describe, with each one very difficult and highly interconnected with the others.

allochka39001 [22]2 years ago

4 0

How the physical world operates.

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A helium balloon has a pressure of 40 psi at 20°C. What will the pressure be at 40°C? Assume constant volume and mass.

Likurg_2 [28]

Answer:

P V = N R T ideal gas equation

P2 / P1 = T2 / T1 where the other variables are constant

P2 = (T2 / T1) * P1 = (313 / 293) * 40 psi = 42.7 psi

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1 year ago

In the desert enviroment the chemical weathering of rocks is generally reduced because

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3 years ago

Light in vacuum travels at a speed of 3.00 x 10^8 m ^-1 s^-1 on average earth is 93,000,000 miles from the sun how many minutes

NemiM [27]

The correct unit for the speed of light is [ m s⁻¹ ].

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Time = (9.3 x 10^7 miles) x (1609 m/mile) / (3 x 10^8 m/s) = 498.8 seconds .

That would be 8.31 minutes.

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2 years ago

If you have 3 equal plates hanging 2 spaces from the center of a balance, at what distance would you need to place 2 equal plate

Reptile [31]

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2 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

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2 years ago