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timama [110]
3 years ago
10

A photon with an energy E = 2.12 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 96.0 MeV. What

is the kinetic energy of the antiproton? Note: m.c2 = 938.3 MeV.
Physics
1 answer:
Oksanka [162]3 years ago
6 0

Answer:

The kinetic energy of the anti proton is 147.4 MeV.

Explanation:

Given that,

Energy = 2.12 GeV

Kinetic energy = 96.0 MeV

We need to calculate the kinetic energy of the anti proton

Using formula of energy

E_{photon}=m_{p}c^2+m_{np}c^2+K.E_{p}+K.E_{np}

We know that,

m_{p}c^2=m_{np}c^2

So, E_{photon}=2mc^2+K.E_{p}+K.E_{np}

K.E_{np}=E_{photon}-(2mc^2+K.E_{p})

Put the value into the formula

K.E_{np}=2.12\times10^{9}-2\times938.3\times10^{6}-96\times10^{6}

K.E_{np}=147.4\ MeV

Hence, The kinetic energy of the anti proton is 147.4 MeV.

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A proton is moving toward a second, stationary proton. What happens as the protons get closer?
MatroZZZ [7]

Answer:

A. Kinetic energy is converted to electric potential energy, and the proton moves more slowly.

Explanation:

When a moving proton is brought close to a stationary one, the kinetic energy of the moving one is converted to electric potential  and the proton moves more slowly.

Kinetic energy is the energy due to the motion of a body. A moving proton will possess this form of energy.

Two protons according to coulombs law will repel each other with an electrostatic force because they both have similar charges. This will increase their electric potential energy of both of them.

Potential energy is the energy at rest of a body. As it increases, the motion of a body will be slower and it will tend towards being stationary.

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3 years ago
Doubling the distance between you and a source of radiation decreases your exposure by:
Elan Coil [88]

Given what we know, we can confirm that doubling the distance between you and a source of radiation decreases your exposure by 75%.

<h3>How is distance related to radiation exposure?</h3>
  • As expected, increasing the distance from the source of the radiation will reduce its negative effects.
  • Counter-intuitively however, doubling the distance does not reduce by half, but rather reduces its effects by 3/4th.
  • This is due to the fact that the radiation effects from the source are inversely proportional to the square of the distance.
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Therefore, given that the radiation is proportional to the square of the distance, instead of being of a more direct relation, we can confirm that when doubling the distance between yourself and the source of the radiation, you can reduce its effects by 3/4 or 75%.

To learn more about radiation visit:

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5 0
3 years ago
A grapefruit falls from a tree and hits the ground .75 seconds later. How far did the grapefruit drop? What was its speed?
Ivanshal [37]
1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration g=9.81 m/s^2. Calling h its height at t=0, the height at time t is given by
h(t)=h- \frac{1}{2}gt^2
We are told thatn when t=0.75 s the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
0=h- \frac{1}{2}gt^2
h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(0.75 s)^2=2.76 m

2) The speed of the grapefruit at time t is given by
v(t)=v_0 +gt
where v_0=0 is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
v(0.75 s)=gt=(9.81 m/s^2)(0.75 s)=7.36 m/s
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Suppose a box is being dragged by a force of 100 n over a rough surface. if the coefficient of kinetic friction at the interface
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I'm sorry if this is wrong but I think it's 2.5 net
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In a perfectly inelastic one-dimensional collision between two objects, what condition alone is necessary so that the final kine
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<span>The condition alone that is necessary so that the final kinetic energy of the system is zero after the collision is that the objects must have momenta with the same magnitude but opposite directions.</span>
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