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3 years ago

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An inflatable backyard swimming pool is being filled from a garden hose with a flowrate of 0.12 gal/s. (a) If the pool is 8 ft i

n diameter, determine the time rate of change of the depth of the water in the pool.

2 answers:

Over [174]3 years ago

7 0

Explanation:

Below is an attachment containing the solution.

zzz [600]3 years ago

6 0

Answer:

0.00318 ft/s

Explanation:

Volume of the inflatables swimming pool, V = πr²h

V = πr²h

a) (dh/dt) = (dh/dV) × (dV/dt)

V = πr²h

(dV/dh) = πr²

(dh/dV) = (1/πr²)

So, we require the rate of change of depth when diameter = 8 ft, radius = (diameter)/2 = 4 ft

(dh/dt) = (dh/dV) × (dV/dt)

(dh/dV) = (1/πr²) = (1/π(4²)) = 0.0199 ft⁻²

(dV/dt) = 0.12 gal/s = 0.016 ft³/s

(dh/dt) = (0.0199 × 0.016) = 0.00318 ft/s

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A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

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Answer:

a

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b

$P_2 = 80600 \ Pa$

Explanation:

From the question we are told that

The pressure of the water in the pipe is $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

Generally the continuity equation is mathematically represented as

$A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity

So

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=> $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$

Here $\rho$ is the density of water with value $\rho = 1000 \ kg /m^3$

$P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]$

=> $P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]$

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3 years ago

Multiply 5.036×102m by 0.078×10−1, taking into account significant figures.

Anika [276]

Answer : The significant digit is 6

Explanation :

Multiply $5.036\times10^{2}m$ by $0.078\times10^{-1}m$

Now, on multiplying

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.392808 \times10^{1}\ m^{2}$

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.0392808\ m^{2}$

Now, the significant digit is 6.

Hence, this is the required solution.

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