We dont see objects. We see the light ____ off objects.

Physics

1 answer:

Romashka [77]2 years ago

4 0

The answer is BBBBBBBBB

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If a ping pong ball and a golf ball are both moving in the same direction with the same amount of kinetic energy, the speed of t

Liono4ka [1.6K]

If the kinetic energy of each ball is equal to that of the other,
then

(1/2) (mass of ppb) (speed of ppb)² = (1/2) (mass of gb) (speed of gb)²

Multiply each side by 2:

(mass of ppb) (speed of ppb)² = (mass of gb) (speed of gb)²

Divide each side by (mass of gb) and by (speed of ppb)² :

(mass of ppb)/(mass of gb) = (speed of gb)²/(speed of ppb)²

Take square root of each side:

√ (ratio of their masses) = ( 1 / ratio of their speeds)²

By trying to do this perfectly rigorously and elegantly, I'm also
using up a lot of space and guaranteeing that nobody will be
able to follow what I have written. Let's just come in from the
cold, and say it the clear, easy way:

If their kinetic energies are equal, then the product of each
mass and its speed² must be the same number.

If one ball has less mass than the other one, then the speed²
of the lighter one must be greater than the speed² of the heavier
one, in order to keep the products equal.

The pingpong ball is moving faster than the golf ball.

The directions of their motions are irrelevant.

5 0

2 years ago

The energies of electrons are said to be quantized. Explain what this means. please explain.

sleet_krkn [62]

To move from one energy level to another, an electron must gain or lose just the right amount of energy. Electrons are said to be quantized because they need a quantum of energy to move to a different sublevel. ... When atoms absorb energy, electrons move into higher energy levels.

8 0

2 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$