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9966 [12]
2 years ago
11

A 5.8 kg bowling ball is rolling down the lane with a velocity of 8.1 m/s . Calculate the momentum of the bowling ball. Provide

your response to the nearest tenth kgm/s .
Physics
1 answer:
zheka24 [161]2 years ago
8 0

Answer:

46.98 kg m/s

Explanation:

P = mv

P = 5.8*8.1 = 46.98 kg m/s

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Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v
wlad13 [49]
1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we callv_f the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f (1)
where
m_A=7 kg is the mass of ball A
m_B=2 kg is the mass of ball B
v_{Ai}=6 m/s is the initial velocity of ball A
v_{Bi}=-12 m/s is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find v_f, we find that the final velocity of the balls is
v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s
and the positive sign means the two balls are going to the right.


2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}
\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2
where
v_{fA} is the final velocity of ball A
v_{fB} is the final velocity of ball B

If we solve simultaneously the two equations, we find:
v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s
v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s
and the total momentum after the collision is:
p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s

3 0
3 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
2 years ago
suppose a vacuum cleaner use 120 j of electrical energy. If 45 j are used to pull air into the vacuum cleaner, how efficient is
Mice21 [21]
It depends on how much time 45 j will go, so if you told me that it went through 45 j per minute it will go through for 2 minutes so not efficient unless it went through 45 j per hour then it is efficient. 
8 0
3 years ago
A time-varying magnetic field is caused by which of the following?
Anit [1.1K]

A time-varying magnetic field can be caused by fluctuating electric fields.

7 0
3 years ago
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