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valentinak56 [21]
3 years ago
6

A light string is wrapped around the edge of the smaller disk, and a 1.50 kg block is suspended from the free end of the string.

If the block is released from rest at a distance of 2.80 m above the floor, what is its speed just before it strikes the floor
Physics
1 answer:
LenaWriter [7]3 years ago
7 0

Answer: 7.41 m/s

Explanation: By using the law of of energy, kinetic energy of the brick as it falls equals the potential energy before falling.

Kinetic energy = mv²/2, potential energy = mgh

mv²/2 = mgh

v²/2 = gh

v² = 2gh

v = √2gh

Where g = 9.8 m/s², h = 2.80m

v = √2×9.8×2.8 = 7.41 m/s

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D 5 V

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Without seeing the whole circuit it is impossible to say for certain.

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A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
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Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

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To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

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Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

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Answer:

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