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3 years ago

How would I answer this equation? E= mc^2 for m

Physics

1 answer:

zubka84 [21]3 years ago

5 0

The E stands for energy. M stands for mass c squared stands for speed of light squared

energy= mass speed of light squared

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A cube with 30-cm-long sides is sitting on the bottom of an aquarium in which the water is one meter deep. (Round your answers t

OleMash [197]

Answer:

A) hydrostatic force on top of cube = 882.9N

B) hydrostatic force on sides of cube = 0N

Explanation:

Detailed explanation and calculation is shown in the image below

4 0

3 years ago

Un alumno menciona que al abrir la ventana de su casa sintió cómo el frío ingresaba a su cuerpo. Menciona cuál es la verdadera r

stepan [7]

Answer:

My believe the answer is

A.) or B.)

Explanation:

Here is why I think A is the answer.

If we use the process of elimination, it would look like this,

a) Porque el aire tiene una temperatura menor que la de su cuerpo; por eso se propaga más rápido.

<em>This makes sense because we all know in winter the weather is very cold and freezing.</em>

b) Porque la temperatura de su cuerpo, siente el aire frio que entra por la ventana.

<em>I feel like this answer is the question, but it could also be an answer, sorry, I'm a little uncertain.</em>

c) Porque el calor de su cuerpo se propaga al medio ambiente, al ser la temperatura del niño mayor que la del aire exterior.

<em>This answer has nothing to do with the question, plus it is very false, our body heat is not enough to overcome the very cold temperature from outside.</em>

d) Porque la temperatura del aire es igual a la temperatura del cuerpo.

<em>This is false because again our body heat is not even compared to the freezing cold temperatures from the winter.</em>

<em />

<h2>Well, have a nice rest of the day!</h2><h3>ba baiii!</h3>

3 0

2 years ago

How do you solve for a net force

AVprozaik [17]

Answer:

Learning the formula.multiply mass accelebrations.the force(F)required to move an object of mass(M) with an acceleration (a) is given by the formula F = m x a.so, force = mass multiplied by accelebration.

8 0

3 years ago

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

2 years ago

What type of force are you exerting when you lie on a bed

zysi [14]

Answer:

Compression is the answer

Explanation:

got it right

6 0

3 years ago