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kiruha [24]
3 years ago
7

A very light ping-pong ball moving east at a speed of 4 m/s collides with a very heavy stationary bowling ball. The Ping-Pong ba

ll bounces back to the west, and the bowling ball moves very slowly to the east. Which object experiences the greater magnitude impulse during the collision?A very light ping-pong ball moving east at a speed of 4 m/s collides with a very heavy stationary bowling ball. The Ping-Pong ball bounces back to the west, and the bowling ball moves very slowly to the east. Which object experiences the greater magnitude impulse during the collision?
Physics
1 answer:
KengaRu [80]3 years ago
3 0

Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

where is the initial momentum of the ping-poll ball

is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

is the final momentum of the ping-poll ball

is the final momentum of the bowling ball

We can re-arrange the equation as follows or

which means (1) so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

(2) therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

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At sunset, red light travels horizontally through the doorway in the western wall of your beach cabin, and you observe the light
Nady [450]

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9.8\cdot 10^{-6}m

Explanation:

For light passing through a single slit, the position of the nth-minimum from the central bright fringe in the diffraction pattern is given by

y=\frac{n \lambda D}{d}

where

\lambda is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

In this problem, we have

\lambda=700 nm = 7.00\cdot 10^{-7}m is the wavelength of the red light

D = 14 m is the distance of the screen from the doorway

d = 1.0 m is the width of the doorway

Substituting n=1 into the equation, we find the distance between the central bright fringe and the first-order dark fringe (the first minimum):

y=\frac{(1)(7.00\cdot 10^{-7} m)(14 m)}{1.0 m}=9.8\cdot 10^{-6}m

6 0
3 years ago
What is the strength of the electric field inside the membrane just before the action potential?
gtnhenbr [62]

Answer:

Incomplete question, check attachment for the graph needed to solve problem.

A 8.1nm........

Explanation:

Electric Field is given as

E=V/d

Where V is voltage

And d is the distance apart

E is the electric field

The voltage V just before action of potential is -70mV,

The value d=8.1nm

d=8.1×10^-9m

E=V/d

E=-70×10^-3/8.1×10^-9

E=-8.6×10^6 N/C

Then the magnitude of the electric field is 8.6×10^6N/C

5 0
3 years ago
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