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o-na [289]
3 years ago
7

As the spaceship travels upward in the sky, some of its kinetic energy will be lost to the universe due to ?

Physics
1 answer:
GaryK [48]3 years ago
6 0

Answer:

Friction !!!

Explanation:

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Balance and beaker of water. The balance will measure mass and the beaker will measure the volume when you take the initial volume without the key subtracting that value from the value with the key in the beaker
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A disc of mass m slides with negligible friction along a flat surface with a velocity v. The disc strikes a wall head-on and bou
qwelly [4]

Answer:

-v/2

Explanation:

Given that:

  • a disc of mass m
  • Collides with the wall going through a sliding motion on on the plane smooth surface.
  • Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.

<u>We know, kinetic energy is given as:</u>

KE_i=\frac{1}{2}. m.v^2

consider this to be the initial kinetic energy of the body.

<u>Now after collision:</u>

KE_f=\frac{1}{4}\times KE_i

KE_f=\frac{1}{4} \times \frac{1}{2}\times m.v^2

Considering that the mass of the body remains constant before and after collision.

KE_f=\frac{1}{2}\times m.(\frac{v}{2})^2

Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.

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3 years ago
Question 5 (1 point) What happens if there is NO oxygen present in cells?
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Answer:

everything would die

Explanation:

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3 years ago
Which example shows energy?
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D because u r using your energy to do all the things

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3 years ago
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Suppose several particles are in concentric circular orbits under the influence of some centripetal component of force, the magn
olga55 [171]

Explanation:

The centripetal force acting on a particle is given by :

F=\dfrac{mv^2}{r}

Since, v=r\omega

F=mr\omega^2

F\propto r\omega^2

Case 1.

If T\propto \sqrt{r}

Since, T=\dfrac{2\pi}{\omega}

So, \omega\propto \dfrac{1}{\sqrt{r} }

\omega^2\propto \dfrac{1}{r}

So, the force becomes,

F\propto r\times \dfrac{1}{r}

F is independent of r

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If T\propto r^{3/2}

\omega\propto \dfrac{1}{r^{3/2}}

\omega^2 \propto \dfrac{1}{r^3}

So, F\propto r\times \dfrac{1}{r^2}

So, the force is inversely proportional to the square of radius.

Case 3.

If T is independent of r, the force will be directly proportional to the radius of orbit.

Hence, this is the required solution.

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