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Novay_Z [31]
3 years ago
9

Water enters a cylindrical tank through two pipes at rates of 250 and 100 gal/min. If the level of the water in the tank remains

constant, calculate the average velocity of the flow leaving the tank through and 8-in inside diameter pipe
Physics
1 answer:
tester [92]3 years ago
3 0

Answer:

The total amount of water that enters the tank is:

250 gal/min + 100 gal/min = 350gal/min.

Then, if the level of the water remains constant, this means that the water leaves the tank at a rate of 350gal/min.

We know that the diameter of the pipe is 8 inches, then the area of the pipe is:

A = pi*(d/2)^2 = 3.14*(4in)^2 = 50.24in^2

now, the flow can be calculated as:

Q = v*A = (velocity*area)

if we want to write our velocity in inches per minute, then we need to write the entering flow in cubic inches:

1 gallon = 231 in^3

then:

350gal/min = (350*231) in^3/min = 80,850 in^3/min.

Then the water that leaves the tank must be the same amoun, we have:

Q = 80,850 in^3/min. = v*A = v*50.24in^2

v =  (80,850in^3/min)/50.24in^2 = 1609.3 in/min.

The velocity of the flow leaving the tank is 1609.3 in/min.

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5 0
3 years ago
In classical physics, consider a 2 kg block hanging on a spring with a spring constant of 50 N/m. Ignore air resistance. The blo
RUDIKE [14]

Answer:

v = 0

Explanation:

This problem can be solved by taking into account:

- The equation for the calculation of the period in a spring-masss system

T = \sqrt{\frac{m}{k} }     ( 1 )

- The equation for the velocity of a simple harmonic motion

x = \frac{2\pi }{T}Asin(\frac{2\pi }{T}t)   ( 2 )

where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block

Hence

T = \sqrt{\frac{2 kg}{50 N/m}} = 0.2 s

and by reeplacing it in ( 2 ):

v = \frac{2\pi }{0.2s}(14cm)sin(\frac{2\pi }{0.2s}(0.9s)) = 140\pi  sin(9\pi ) = 0

In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.

5 0
3 years ago
Sino tao ang gutom?<br>A.you<br>B.mama mo!<br>C.tamod​
Fiesta28 [93]
The answer to that question is c. tamod
5 0
3 years ago
What is the intensity of 60dB sound?​
polet [3.4K]

Answer:

The intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

Explanation:

Given;

intensity of the sound level, dB = 60 dB

The intensity of the sound in W/m² is calculated as;

dB = 10 Log[\frac{I}{I_o} ]\\\\

where;

I₀ is threshold of hearing = 1 x 10⁻¹² W/m²

I is intensity of the sound in W/m²

Substitute the given values and for I;

dB = 10 Log[\frac{I}{I_o} ]\\\\60 = 10 Log[\frac{I}{I_o} ]\\\\6 =  Log[\frac{I}{I_o} ]\\\\10^6 = \frac{I}{I_o} \\\\I = 10^6 \ \times \ I_o\\\\I = 10^6 \ \times \ 1^{-12} \ W/m^2 \\\\I = 1\ \times \ 10^{-6} \ W/m^2

Therefore, the intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

5 0
3 years ago
Light of wavelength 559 nm is used to illuminate normally two glass plates 22.1 cm in length that touch at one end and are separ
umka21 [38]

Answer:

M = 222 fringes

Explanation:

given

λ = 559 n m = 559 × 10⁻⁹ m

radius = 0.026 mm = 0.026 ×10⁻³ m

length of the glass plate = 22.1 ×10⁻² m

using relation

2t=(m+\dfrac{1}{2})\lambda\ \ (m=0,1,2,3...)\\where\ 0\leq t\leq 2r\\m = \dfrac{2t}{\lambda}-\dfrac{1}{2}

m_{max} = \dfrac{2\times 2r}{\lambda}-\dfrac{1}{2}\\m_{max} = \dfrac{2\times 2\times 0.026\times 10^{-3}}{559\times 10^{-9}}-\dfrac{1}{2}

 = 221.79  

 = 221 (approx.)

hence no of bright fringe

M = m + 1

   = 221 +1

M = 222 fringes

6 0
3 years ago
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