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schepotkina [342]
3 years ago
5

A billiard ball of mass m hits another one of the same mass. The first ball moves off at 30 degrees. For an elastic collision wh

at are the velocities of both balls after the collision. In the lab frame where the second ball starts at rest, what angle compared to the direction of initial velocity of the first ball does the 2nd ball move?
Physics
1 answer:
sleet_krkn [62]3 years ago
6 0

Answer:

v₁ = u₁/2√3 ≈ 0.866u₁

v₂ = u₁/2      = 0.5u₁

θ = 60°

Explanation:

let u₁ be the initial velocity of the first ball

let v₁ be the final velocity of the first ball

let v₂ be the final velocity of the second ball

For elastic collisions, the angle between the departing masses is 90°

assume the first ball initially moves along the x axis in the positive direction

conservation of momentum

In the y direction, initial momentum is zero

After the collision

mv₁sin30 = mv₂sin60

½v₁ = ½√(3)v₂

v₁ = √(3)v₂

in the x direction,

mu₁ = mv₁cos30 + mv₂cos-60

u₁ = v₁cos30 + v₂cos60

u₁ = (√(3)v₂)½√(3) + ½v₂

u₁ = 2v₂

v₂ = u₁/2

v₁ = √(3)v₂ = √(3)(u₁/2)

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daser333 [38]

Answer:

The magnitude of force is 1.86 N and the direction of force is towards the other wire.

Explanation:

Given:

Current flowing through each power line, I = 130 A

Distance between the two power lines, d = 40 cm = 0.4 m

Length of power lines, L = 220 m

The force exerted by the power lines on each other is given by the relation:

F = \frac{\mu_{0}LII }{2\pi d}

Substitute the suitable values in the above equation.

F = \frac{4\pi\times10^{-7}\times220\times130\times130 }{2\pi\times0.4}

F = 1.86 N

Since the direction of current flowing through the power lines are opposite to each other, so the force is attractive in nature. Hence, the direction of force experienced by the power lines on each other is towards the each other.

5 0
3 years ago
A 1250 kg car is stopped at a traffic light. A 3550 kg truck moving at 8.33 m/s hits the car from behind. If bumpers lock, how f
Radda [10]

Answer:

the two vehicles will be moving at a speed of 6.16  m/s

Explanation:

This is a case of completely inelastic collision, therefore, the conservation of momentum can be written as:

m_1\,v_1+m_2\,v_2=(m_1+m_2)\,v_f

which given the information provided results into:

m_1\,v_1+m_2\,v_2=(m_1+m_2)\,v_f\\(1250)\,(0)+(3550)\,(8.33)=(1250+3550)\,v_f\\29571.5=4800\,v_f\\v_f=6.16\,\,m/s

7 0
3 years ago
Show all work please I am stuck​
MissTica

Answer:

Explanation:

1 meter = 39.37 inches

meters x inches/meters = inches

6.23x10^-4 m x 39.37 in./m=0.245...=2.45x10^-2

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3 years ago
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Vera_Pavlovna [14]

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5 0
2 years ago
An electron with charge −e and mass m moves in a circular orbit of radius r around a nucleus of charge Ze, where Z is the atomic
shepuryov [24]

Answer:

v=\sqrt{\frac{kZe^2}{mr}}

Explanation:

The electrostatic attraction between the nucleus and the electron is given by:

F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2} (1)

where

k is the Coulomb's constant

Ze is the charge of the nucleus

e is the charge of the electron

r is the distance between the electron and the nucleus

This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:

F=m\frac{v^2}{r} (2)

where

m is the mass of the electron

v is the speed of the electron

Combining the two equations (1) and (2), we find

k\frac{Ze^2}{r^2}=m\frac{v^2}{r}

And solving for v, we find an expression for the speed of the electron:

v=\sqrt{\frac{kZe^2}{mr}}

6 0
3 years ago
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