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coldgirl [10]
2 years ago
15

Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the sys

tem of two particles assosicated with the force is given by the equation U=G/r 2, where r is the distance between the two particles and G is a positive constant
Physics
2 answers:
Archy [21]2 years ago
5 0

Speed of particle B is 2v₀/3 m/s to the left. Particle A and particle B will always have equal speed since they experience equal forces.

<h3>Conservation of energy</h3>

The speed and direction of the particle B is determined by applying the principle of conservation of energy as follows;

K.E₁ + P.E₁ = K.E₂ + P.E₂

\frac{1}{2} Mv^2_A + \frac{G}{r^2} = \frac{1}{2} Mv^2_B + \frac{G}{r^2} \\\\ \frac{1}{2} Mv^2_A = \frac{1}{2} Mv^2_B\\\\v^2_A = v^2_B\\\\v_A = v_B

v_B = \frac{2v_0}{3}  \ m/s \ to \ the \ left

At any given position, the speed of particle A and particle B will be equal, since they experience equal force and they have equal masses.

The complete question is below:

Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the system of two particles assosicated with the force is given by the equation U=G/r 2, where r is the distance between the two particles and G is a positive constant. At time t=T1 particle A is observed to be traveling with speed 2vo/3 to the left. The speed and direction of motion of particle B is ?

Learn more about conservation of energy here: brainly.com/question/166559

Trava [24]2 years ago
3 0

Answer:

Explanation:

the linear momentum is conserved MiVi=MaVa+MbVb

MV0=2V0M/3+MVb

Vb=V0/3

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1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

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diameter = 0.550 cm

now if the current in the ammeter is given as

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now we know that

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Part b)

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2.

Answer:

Part a)

i = 1.55 A

Part b)

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Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

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Part b)

now we have

j = nev_d

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n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

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