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2 years ago

Water enters a shower head through a pipe of radius 0.0112 m at 3.25 m/s. What is it’s volume flow rate? (Unit= m^3/s)

Physics

2 answers:

kenny6666 [7]2 years ago

7 0

Answer:

$1.28 \times 10 ^{-3} m^3/s$

Explanation:

Given that the radius of the pipe, r =0.0112 m

So, the area of the cross-section, $a= \pi (0.0112)^2 = 3.941\times 10^{-4} m^2$

Speed of water in the pipe, v=3.25 m/s

Volume flow rate = $av= 3.941\times 10^{-4} \times 3.25 = 1.28\times 10 ^{-3} m^3/s$

Hence, the volume flow rate is $1.28 \times 10 ^{-3} m^3/s$

tangare [24]2 years ago

4 0

Answer:

For Acellus the answer is 0.00128

Explanation:

Trust me

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2 years ago

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A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

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Answer:

1.2 m/s

0.31 m

0.15 m

Explanation:

Time period is

$T=2.5\times 2\\\Rightarrow T=5\ s$

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$f=\dfrac{1}{T}\\\Rightarrow f=\dfrac{1}{5}\\\Rightarrow f=0.2\ Hz$

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$v=f\lambda\\\Rightarrow v=0.2\times 6\\\Rightarrow v=1.2\ m/s$

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Amplitude is given by

$A=\dfrac{d}{2}\\\Rightarrow A=\dfrac{0.62}{2}\\\Rightarrow A=0.31\ m$

Amplitude is 0.31 m

If d = 0.3 m

$A=\dfrac{0.3}{2}=0.15\ m$

The amplitude would be 0.15 m. The speed would remain the same.

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3 years ago

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2 years ago

Abnormal protrusion of the eye out of the orbit is known as

kifflom [539]

Answer:

Exophthalmos

Explanation:

Exophthalmos is a disorder which can be either bilateral or unilateral. Sometimes it is also known by other names like Exophthalmus, Excophthamia, Exobitism.

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2 years ago

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words: