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topjm [15]
3 years ago
6

A 9.0 kg bowling ball races down the lane at 15 m/s before striking a bowling pin (at rest) with a mass of 0.85 kg. If the 0.85

kg pin bounces backward with a velocity of 45 m/s, what is the velocity of the bowling ball after the collision
Physics
2 answers:
artcher [175]3 years ago
8 0

Answer:

v = 10.75\,\frac{m}{s}

Explanation:

The system ball-pin is modelled by the Principle of Moment Conservation:

(9\,kg)\cdot (15\,\frac{m}{s} ) + (0.85\,kg)\cdot (0\frac{m}{s} ) = (9\,kg)\cdot v + (0.85\,kg)\cdot (45\,\frac{m}{s} )

The velocity of the bowling ball after the collision is:

v = 10.75\,\frac{m}{s}

Karolina [17]3 years ago
5 0

Answer:

v1 = 10.75m/s

Explanation:

Given m = 9.0kg

u1 = 15m/s, m2 = 0.85kg, u2 = 0m/s bowling pin initially at rest. v1 =? v2 = 45m/s

Where u represents initial velocity and v final velocity for the bodies involved in the collision.

From the principle of conservation of momentum, the sum of momentum before collision is equal the sum of momentum after collision.

m1u1 + m2u2 = m1v1 + m2v2

9×15 + 0.85×0 = 9v1 + 0.85×45

135 = 9v1 +38.25

9v1 = 135 – 38.25

9v1 = 96.75

v1 = 96.75/9 = 10.75m/s

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Answer:

it is reduced four times.

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3 years ago
The amplitude of a standing sound wave in a long pipe closed at the left end is sketched below.
ollegr [7]

(1) The harmonic number for the mode of oscillation is 3.

(2) The pitch (frequency) of the sound is 579.55 Hz

(3) The level of the water inside the vertical pipe is 0.1 m.

<h3>The harmonic number</h3>

The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.

<h3>Frequency of the wave</h3>

The pitch (frequency) of the sound is calculated from third harmonic formula;

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Learn more about harmonics of closed pipes here: brainly.com/question/27248821

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1 year ago
after singing for a few minutes the soloist rejoins the choir a second choir, consist of 90 additional singers then joins in eac
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