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docker41 [41]
3 years ago
7

Alright eksqijakojqnlqozjzbw.wlisj

Physics
1 answer:
saw5 [17]3 years ago
6 0

Answer:

eksqijakojqnozjzbw.wlisjaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Explanation:

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In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
On a roller coaster ride the total mass of a cart - with two passengers included - is 319 kg. Peak K is at 43.6 m above the grou
Dafna1 [17]

The mechanical energy is lost due to friction between the two peak is 78,458.688 J

<h3>What is mechanical energy?</h3>

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a roller coaster ride the total mass of a cart - with two passengers included - is 319 kg. Peak K is at 43.6 m above the ground and peak L is at 24.4 m. At location K the speed of the cart is 16.4 m/s, and at location L it is 12.4 m/s.

Total energy at peak K,

TE₁ = 1/2 mv₁² +mgh₁

Substitute the values, we get

TE₁ =  1/2 x319 x 16.4² +319 x 9.81 x 43.6

TE₁ = 179,340. 524 J

Total energy at peak L,

TE₂ = 1/2 mv₂² +mgh₂

Substitute the values, we get

TE₂ =  1/2 x319 x 12.4² +319 x 9.81 x 24.4

TE₂ = 100,881. 836 J

The mechanical energy lost is

M.E =  TE₁ -TE₂

M.E =  179,340. 524 J -  100,881. 836 J

M.E = 78,458.688 J

Thus, the mechanical energy is lost due to friction between the two peak i

Learn more about mechanical energy.

brainly.com/question/13552918

#SPJ1

7 0
2 years ago
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