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Rus_ich [418]
3 years ago
14

A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

ight line so that the two balls have the same momentum?
Answer in units of m/s.
Physics
1 answer:
lilavasa [31]3 years ago
6 0

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

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Where,

F – Force, G – gravitational constant, M and m – masses in kg, r – distance in meters.

Since force is proportional to the masses of interacting objects. If the mass of any one object increases, gravity between them also gets increased. When moving to higher altitude, force decreases as the distance is inverse proportion to gravity.

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A meter stick flies by with an apparent length of 60 cm. What is it's velocity in m/s? Not sure which formula to use
vichka [17]

Solution:

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A gold nucleus is 466 fm (1 fm = 10-15 m) from a proton, which initially is at rest. When the proton is released, it speeds away
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Answer:

v=6.8\times10^6m/s

Explanation:

The sum of the kinetic and electric potential energies of the proton when initially released must be equal to their sum at infinity, so we have:

K_i+U_i=K_f+U_f

Which, since K_i=0J because initially the proton is at rest, is:

\frac{kqQ}{d}=\frac{mv^2}{2}+\frac{kqQ}{r_\infty}

where k=9\times10^9Nm^2/C^2 is Coulomb's constant, q=1.6\times10^{-19}C the charge of the proton, Q=(79)(1.6\times10^{-19})C the charge of the gold nucleus, since it has 79 protons, d=466\times10^{-15}m the initial separation between them, m=1.67\times10^{-27}kg the mass of the proton and v its final velocity. r_\infty is very far away, so the final electric potential will be 0J, and we have:

v=\sqrt{\frac{2kqQ}{md}}=\sqrt{\frac{2(9\times10^9)(1.6\times10^{-19})^2(79)}{(1.67\times10^{-27})(466\times10^{-15})}}m/s=6839409m/s

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1 A steel ball of mass 5 kg is released from rest at a height of 3 m above the ground. It rolls down the frictionless section AB
KonstantinChe [14]

The speed of the ball when it reached the point B is 7.7 m/s.  The angle of inclination of section BC is  11.54⁰.

<h3>What is frictional force?</h3>

The force between the two mating surfaces and having the relative motion is called the frictional force.

The speed of ball at point B is equal to the relation

v  = √(2gh)

Substitute the values of height h = 3m and acceleration due to gravity g =9.8 m/s², we get the speed at B

v =√(2 x 9.8 x 3)

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Thus the speed of ball when it reaches the point B is 7.7 m/s.

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The frictional force is equal to the product of coefficient of friction and the normal force on the steel ball \.

f = μN

f =μmg

Substitute the values into the equation, we get the coefficient of friction.

10 = μ x 5 x 9.8

μ = 0.2041

Angle of inclination for the section BC is

θ = tan⁻¹(0.2041) = 11.54⁰

Thus, the angle of inclination of section BC is  11.54⁰.

Learn more about frictional force.

brainly.com/question/14662717

#SPJ1

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