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2 years ago

12

8. A 12kg bowling ball has a velocity of 8 m/s, and is brought to

1 answer:

Natali5045456 [20]2 years ago

8 0

Answer:

The change in momentum experienced by the bowling ball is 96 kgm/s.

Explanation:

Given;

mass of the bowling ball, m = 12 kg

velocity of the bowling ball, v = 8 m/s

time of motion, t = 10 s

The change in momentum experienced by the bowling ball is equal to the impulse experienced by the bowling ball.

ΔP = J = F x t

$J = ft =\frac{mv}{t} \times t = mv\\\\J = 12 \times 8\\\\J = 96 \ kg.m/s$

Therefore, the change in momentum experienced by the bowling ball is 96 kgm/s.

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Essam is abseiling down a steep cliff. How much gravitational potential energy does he lose for every metre he descends? His mas

Dafna11 [192]

Answer:

720 J

Explanation:

The gravitational potential energy that Essam loses for every metre is given by:

$\Delta U=mg \Delta h$

where

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A closed-end organ pipe is used to produce a mixture of sounds. The third and fifth harmonics in the mixture have frequencies of

nexus9112 [7]

Answer:

$F_1=366.67Hz$

Explanation:

From the question we are told that:

Frequency of 3rd harmonics $F_3=1100$

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Generally the equation for Wavelength at 3rd Harmonics is mathematically given by

$\lambda_3=\frac{4}{3}l$

Therefore

$F_3=\frac{3v}{4l}$

Generally the equation for Wavelength at 1st Harmonics is mathematically given by

$\lambda_1=\frac{4}{1}l$

Therefore

$F_1=\frac{v}{4l}$

Generally the equation for the frequency of the first harmonic is mathematically given by

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A train travels 600km in 4 hours. What is the speed of the train?

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Answer: 150 miles per hour

Explanation:

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Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pull

bearhunter [10]

Answer:

$strain = 1.4 \times 10^{-3}$

Explanation:

As we know by the formula of elasticity that

$E = \frac{stress}{strain}$

now we have

$E = 110 GPA$

$E = 110 \times 10^9 Pa$

Area = 15.2 mm x 19.1 mm

$A = 290.3 \times 10^{-6}$

now we also know that force is given as

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here we have

stress = Force / Area

$stress = \frac{44500}{290.3 \times 10^{-6}}$

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now from above formula we have

$strain = \frac{stress}{E}$

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