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2 years ago

WILL VOTE MOST BRAINLIEST

Physics

1 answer:

andrezito [222]2 years ago

8 0

Answer:

B) The lighter gases boiled off of the protoplanets closest to the sun, leaving dust and metals behind to form the inner planets.

Explanation:

Inner planets are smaller and rockier than outer gas planets,outer planets are larger,because of their lower gravity they don't attract extreme amounts of gas in their planets.The four outer planets were so far from the Sun that its winds could not blow away their ice and gases

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If you fired a rifle straight upwards at 1000 m/s, how far up will the bullet get?

nadya68 [22]

Answer:

h = 51020.40 meters

Explanation:

Speed of the rifle, v = 1000 m/s

Let h is the height gained by the bullet. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

$h=\dfrac{v^2}{2g}$

$h=\dfrac{(1000\ m/s)^2}{2\times 9.8\ m/s^2}$

h = 51020.40 meters

So, the bullet will get up to a height of 51020.40 meters. Hence, this is the required solution.

4 0

2 years ago

How is temperature related to kinetic energy

stiks02 [169]

Temperature is the average kinetic energy of an object

7 0

3 years ago

An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e

timama [110]

Answer:

K = 80.75 MeV

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

$E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}$

<em>where $E_{ph}$ : is the photon energy, $E_{0p} and E_{0ap}$ : are the rest energies of the proton and the antiproton, respectively, equals to m₀c², $K_{p} and K_{ap}$ : are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>

Therefore the kinetic energy of the antiproton is:

$K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}$

<u>The proton mass is equal to the antiproton mass, so</u>:

$K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV$

$K_{ap} = 80.75 MeV$

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

3 0

3 years ago

A positively charged particle moves through an electric field. As part of a complicated trajectory, the particle passes through

kow [346]

Answer:

(B) The speed is larger at A than at B.

Explanation:

Point B, the final point of the trajectory, has higher electric potential than point A, the initial point of the trajectory, so the electric potential energy of the charged particle increases, which means that its kinetic energy must be decreasing, thus the speed at B must be lower than the speed at A.

8 0

3 years ago

Linus builds an electrical circuit with a battery and with wires that carry the current. As the battery weakens, the current als

pantera1 [17]

Answer:

The field gets weaker

Explanation:

I’m taking the test right now, hope this helps!!

7 0

3 years ago

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