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3 years ago

Who created E = mc2? (What is it)?

1 answer:

5 0

Albert Einstein created this equation. It means that the formula $E= mc^{2}$ shows that mass can be converted into energy, and energy can be converted into mass. Einstein made this equation 100+ years ago, and yes, he was right with the theory.

(Mark this as the brainliest if this answer helps!)

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PLEASE SAY IM RIGHT AGAIN?

zhuklara [117]

I believe it’s the scientific revolution
Hope this helps :)

3 0

3 years ago

A bag of cement weighing 325 N hangs in equilibrium from three wires. Two of the wires make angles of theta1=60.0 degrees and th

Murljashka [212]

The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement.

The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N

The other free body diagram is around the joint of the three strings.

In this case, you can do the horizontal forces equilibrium equation as:

T1* cos(60) - T2*cos(40) = 0

And the vertical forces equilibrium equation:

Ti sin(60) + T2 sin(40) = T3 = 325 N

Then you have two equations with two unknown variables, T1 and T2

0.5 T1 - 0.766 T2 = 0

0.866 T1 + 0.643T2 = 325

When you solve it you get, T1 = 252.8 N and T2 = 165 N

Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N

7 0

3 years ago

Si tu velocidad al caminar es 0,3 m/s. Determina el tiempo que demorarias en llegar al sol en horas (REALIZAR TRANSFORMACIÓN DE

Contact [7]

Answer:

138,516,546.9 horas.

Explanation:

Tenemos que usar la ecuación:

Velocidad = distancia/tiempo

Acá tenemos:

Velocidad = 0.3m/s

distancia = 149597870700 m

y queremos resolver la ecuación para el tiempo:

0.3m/s = 149597870700m/tiempo.

tiempo = 149597870700m/(0.3m/s) = 498,659,569,000 s

y sabemos que una hora tiene 3600 segundos, entonces si queremos transformar de segundos a horas tenemos:

498,659,569,000 s = (498,659,569,000/3600) h = 138,516,546.9 horas.

6 0

3 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$