All categories

4 years ago

Who does the little boy in the animation with the compass grow up to be??

Physics

1 answer:

sattari [20]4 years ago

5 0

Albert Einstein does the little boy in the animation with the compass grow up to be.

<u>Explanation:</u>

Albert Einstein, was raised at his family home in Germany, one of the finest thinkers of all time in science,on March 14, 1879 he celebrates his birthday with Pi Day, a remembrance of the special number that exist indefinitely. He did very well in his classrooms considering some arguments, particularly math and science, but he did not liked the teaching methodologies during his school time. His father's offering of a compass when he was five years old contributed to an everlasting passion with magnetic fields. This led him to highlight his name in the world forever.

You might be interested in

Calculate the potential energy of a 4 kg cat crouched 3 meters off the ground

Tom [10]

Gravitational potential energy = mass × gravity × height

Ep = (4)(9.81)(3)

Energy = 117.72 Joules

= 1.2x10^2 Joules

7 0

3 years ago

Read 2 more answers

A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

Convert $24.6\; \rm \text{hours}$ to seconds:

$24.6 \times 3600 = 88560\; \rm s$

Solve the equation for $R$ :

$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

5 0

4 years ago

Help me pls???!!

rosijanka [135]

Answer:

A. I = V / R = 12 / 252 = .048 amps

V = I * R = .048 * 252 = 12 V

V is also the reading the voltage across the battery (12 Volts)

4 0

3 years ago

If a wave has a wavelength of 9 meters and a period of 0.006, what is the velocity of the wave?

lana66690 [7]

<u>Answer</u>

D. 1,500 m/s

<u>Explanation</u>

the wave equation states that,

V = λf

Where V ⇒ Velocity

λ ⇒ wavelength

f ⇒ Frequency

F = 1/T

Where T ⇒ period

F = 1/0.006

= 166.667

∴ V = 9 × 166.667

= 1,500 m/s

4 0

3 years ago

Read 2 more answers

Explain how that energy converts itself from POTENTIAL to KINETIC energy.

Karolina [17]

Answer:

Potential energy can transfer into other forms of energy like kinetic energy. Kinetic energy is energy an object has because of its motion.

Explanation:

The ball was released, as the ball moves faster and faster toward the ground, the force of gravity will transfer the potential energy to kinetic energy.

8 0

3 years ago