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kap26 [50]
2 years ago
7

A manometer is used to measure the air pressure in a tank. the fluid used has a specific gravity of 1.25, and the differential h

eight between the two arms of the manometer is 28 in. if the local atmospheric pressure is 12.7 psia, determine the absolute pressure in the tank for the cases of the manometer arm with the (a) higher and (b) lower fluid level being attached to the tank.
Physics
1 answer:
BartSMP [9]2 years ago
4 0
Specific Gravity of the fluid = 1.25 
Height h = 28 in
 Atmospheric Pressure = 12.7 psia
 Density of water = 62.4 lbm/ft^3 at 32F
 Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
 Density of the Fluid p = 78 lbm/ft^3
 Difference in pressure as we got the differential height, dP = p x g x h  dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
 Difference in pressure = 1.26 psia
 (a) Pressure in the arm that is at Higher 
 P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
 (b) Pressure in the tank that is at Lower
 P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
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Answer:

The current needed to transmit Power of 4 W is 28.47 A

Solution:

As per the question:

Length of the antenna, L_{a} = 1 m

Frequency, \vartheta = 1.5 MHz = 1.5\times 10^{6} Hz

Power transmitted, P_{t} = 4 W

Now,

For a monopole antenna:

\lambda_{a} = \frac{c}{\vartheta}

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\lambda_{a} = wavelength transmitted by the antenna

c = speed of light in vacuum

\lambda_{a} = \frac{3\times 10^{8}}{1.5\times 10^{6}} = 200 m

Now,

Since, the value of \lambda_{a} >> L_{a} thus the monopole is a Hertian monopole.

The resistance is calculated as:

R = \frac{1}{2}(\frac{dL_{a}}{\lambda_{a}})^{2}\times 80\pi^{2}

R = \frac{1}{2}(\frac{1}{200)^{2}\times 80\pi^{2} = 9.869\times 10^{- 3} = 9.869 m\Omega

P_{radiated} = P_{t}

P_{radiated} = \frac{R}{I^{2}}

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2 years ago
A ball on a string travels once around a circle with a circumference of 2.0 m. The tension in the string is 5.0 N. how much work
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Answer:0

Explanation:

Given

circumference of circle is 2 m

Tension in the string T=5 N

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In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is 90^{\circ}

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Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

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Therefore approximately, Fa=774 N and Fb=346 N

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