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kow [346]
3 years ago
9

Two pipes move the same amount of ideal fluid in the same amount of time. One pipe has a 2 in. diameter; the other has a 3 in. d

iameter. In which pipe is the fluid pressure higher, assuming that all other conditions are the same?a) 3-in. pipe b) 2-in. pipe c) same pressure in both d) cannot be determined for an ideal fluid
Physics
1 answer:
KATRIN_1 [288]3 years ago
6 0

Answer:

a) 3-in. pipe

Explanation:

Given that

Fluid flow is in same amount in the same time it means that volume flow rate is same for the pipes

Volume flow rate

Q = A V

A=Area ,V=Velocity

A=\dfrac{\pi}{4}d^2

If diameter d is more then the velocity will be less for same volume flow rate .We also Know that if pressure is more then the velocity will be less.

The second pipe 3 in diameter having more diameter then the velocity will be less but the pressure will be more.

That is why the 3 in diameter is having more pressure than 2 in diameter pipe.

Therefore the answer will be a.

a) 3-in diameter  pipe

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If no part is submerged (V = 0) that is volume. Therefore there is Zero Buoyant Force.

Fully submerged produces greatest buoyant force since greatest amount of fluid was displaced.

Whenever it is fully submerged it will have the greatest buoyant force.

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2 1 3

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7 0
2 years ago
Consider the following three statements: (i) For any electro-magnetic radiation, the product of the wavelength and the frequency
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Answer:

A and B

Explanation:

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c=frequency\times Wavelength

c is the speed of light having value 3\times 10^8\ m/s

Thus, the product of the wavelength and the frequency is constant and equal to 3\times 10^8\ m/s

<u>Option A is correct.</u>

Given, Frequency = 1\times 10^{18}\ Hz

Thus, Wavelength is:

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<u>Wavelength = 3.0 Å</u>

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3 0
3 years ago
Read 2 more answers
Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

3 0
3 years ago
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