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madam [21]
3 years ago
12

A billiard ball moves with 3 kg⋅m/s of momentum and strikes three other billiard balls. What is the total momentum of the balls

after the collision?
(1 point)
more than 3 kg⋅m/s
3 kg⋅m/s
It depends on the mass of each ball.
less than 3 kg⋅m/s
Physics
1 answer:
maxonik [38]3 years ago
3 0

A billiard ball moves with 3 kg⋅m/s of momentum and strikes three other billiard balls that have been just sitting there at rest and not moving.

The total momentum of all four balls after the collision is <em>3 kg⋅m/s</em>, because momentum is not created or destroyed.  The total amount of it after an event is the same as the total amount of it before the event.

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Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
dezoksy [38]
The De Broglie wavelength of the electron is
\lambda=34.0 nm=34 \cdot 10^{-9} m
And we can use De Broglie's relationship to find its momentum:
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s

Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}=  2.13 \cdot 10^4 m/s

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J
3 0
3 years ago
A ball of mass 0.500 kg is carefully balanced on a shelf that is 2.70 m above the ground. What is its gravitational potential en
ratelena [41]

Answer:

The gravitational potential energy of the ball is 13.23 J.

Explanation:

Given;

mass of the ball, m = 0.5 kg

height of the shelf, h = 2.7 m

The gravitational potential energy is given by;

P.E = mgh

where;

m is mass of the ball

g is acceleration due to gravity = 9.8 m/s²

h is height of the ball

Substitute the givens and solve for gravitational potential energy;

PE = (0.5 x 9.8 x 2.7)

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8 0
3 years ago
2) Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another
worty [1.4K]

Answer:

m_1 / m_2 = sqrt (1 / 2)

Explanation:

Given:

- Initial velocity of both skaters V_i = 0

- Velocity of skater 1 after push = V_1

- Velocity of skater  after push = V_2

- Distance traveled by skater 1 = s_1

- Distance traveled by skater 2 = s_2

- s_1 = 2*s_2

- Accelerations of both skaters to halt is equal

Find:

What is the ratio m1/m2 of their masses

Solution:

- Apply conservation of momentum for two skaters just before and after the push as follows:

                                              P_i = P_f

                                  0 = m_1*V_1 - m_2*V_2

- Evaluate:                 m_1 / m_2 = ( V_2 / V_1 )

- Apply Conservation of Energy on both skaters as follows:

- Skater 1:

                               0.5*m_1*V_1^2 = u_k*m_1*g*s_1

-Simplify:                      0.5*V_1^2 = u_k*g*(2*s_2)

- Skater 2:

                               0.5*m_2*V_2^2 = u_k*m_2*g*s_2

-Simplify:                      0.5*V_2^2 = u_k*g*s_2

- Divide the two energy equations for skaters:

                                    (V_1 / V_2)^2 = 2

                                    (V_2 / V_1)^2 = 1 / 2

- simplify:                     (V_2 / V_1) = sqrt (1 / 2)

-Hence from earlier momentum conservation results:

                                  m_1 / m_2 = ( V_2 / V_1 ) = sqrt (1 / 2)

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RoseWind [281]

Answer:

4n

Explanation:

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