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BARSIC [14]
3 years ago
8

A bowling ball moving with a velocity of 5V to the right collides elastically with a beach ball moving at a velocity 2V to the l

eft. The bowling ball barely slows down. What is the approximate velocity of the beach ball after the collision?
Physics
1 answer:
katen-ka-za [31]3 years ago
3 0

Answer:

v'_2=3V

Explanation:

From the question we are told that:

Bowling ball Speed v_1=5 m/s

Beach ball Speed v_2=2 m/s

Let The Mass be equal i.e

 M_1=M_2

Therefore

Generally the equation for Velocity of beach ball after collision v'_2 is mathematically given by

Since Velocity is Vector Quantity

Therefore

 v'_2=v_1-v_2

 v'_2=5-2

 v'_2=3V

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The first trophic level of the food chain has the most energy.

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Which statements describe velocity and acceleration? Check all that apply.
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4 0
3 years ago
Read 2 more answers
Calculate the internal axial load at a point D if length L=7 ft. The part is subjected to loads P1=632 lbs, P2=888 lbs (applied
liraira [26]

Answer:

- 256 lbs

Explanation:

The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero

EF_x = 0; Then:

internal axial load at point D = Δ P

= -(P₂ - P₁)

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5 0
3 years ago
A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

\theta = \omega t\\\\\theta = (100 \frac{rev}{\min}  \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi  \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

6 0
3 years ago
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