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Varvara68 [4.7K]
3 years ago
10

Who created the theory of general relativity?

Physics
2 answers:
Marianna [84]3 years ago
8 0
Albert eisten should be the anwser
pantera1 [17]3 years ago
7 0
Who created the theory of general relativity?:

The answer would be: Albert Einstein.

Albert Einstein developed the general relativity theory (gravitation). 

The year he developed the general relativity theory or (GR) was back in 1907 and 1915. Then many other contributions after 1915.

Thanks,

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A 0.09-kg lead bullet traveling 182 m/s strikes an armor plate and comes to a stop. If all of the bullet's energy is converted t
Artemon [7]

Answer: Δθ = 127.4 K

Explanation: by using the law of conservation of energy, the kinetic energy of the bullet equals the heat energy on the plate.

Kinetic energy of bullet = mv²/2

Heat energy = mcΔθ

Where m = mass of bullet = 0.09kg, v = velocity of bullet = 182 m/s, c = specific heat capacity of lead bullet = 130 j/kgk

Δθ = change in temperature

mv²/2 = mcΔθ

With 'm' on both sides of the equation, they cancel out each other, hence we have that

v²/2 = cΔθ

v² = 2cΔθ

Δθ= v²/2c

Δθ = (182)²/2×130

Δθ = 33124/260

Δθ = 127.4 K

3 0
3 years ago
In a chemical reaction blank are the substances left over
Zepler [3.9K]

Answer: In a chemical reaction blank are the substances left over

Explanation: The substance left over after a reaction takes place

3 0
3 years ago
How do you find the voltage of a section of a parallel circuit?
sergiy2304 [10]

Answer: Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source. You can find total resistance in a Parallel circuit with the following formula: 1/Rt = 1/R1 + 1/R2 + 1/R3 +.

Hope this helps!

3 0
3 years ago
A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the
Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge 15\times10^-6 C is placed which is the origin.

Let B be the point where the charge 9\times 10^-6 C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, \frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2},

Q_A = 15\times 10^-6 C

Q_B=9\times10^-6C

d_A = 1+x cm

d_B=x cm

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

#SPJ4

3 0
1 year ago
What is the acceleration experiance by a car that takes 10s to reach 27m/s from rest
lina2011 [118]

Magnitude of acceleration = (change in speed) / (time for the change).

Change in speed  =  (27 - 0)  =  27 m/s
Time for the change = 10 s

Magnitude of acceleration =  (27 m/s) / (10 s)  =  2.7 m/s²  .
4 0
3 years ago
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