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Vlada [557]
3 years ago
15

A vacuum gage attached to a power plant condenser gives a reading of 27.86 in. of mercury. The surrounding atmospheric pressure

is 14.66 lbf/in. Determine the absolute pressure inside the condenser, in lbf/in. The density of mercury is 848 lb/ft and the acceleration of gravity is g = 32.0 ft/s?.
Physics
1 answer:
Nadusha1986 [10]3 years ago
4 0

Answer:

absolute pressure =  1.07 lbft/in^2

Explanation:

given data:

vaccum gauge reading h = 27.86 inch = 2.32 ft

we know that

gauge pressure p is given as

p = \rho gh

p = 848 \ lb/ft^3 * 32 \ ft/s^2 *2.32 ft  = 62955.52 \ lbft/s^2 * 1/ft^2

we know that  1\  lb ft\s^2 = \frac{1}{32.174}\  lbft

1 ft = 12 inch

therefore p = 62955.52 * \frac{1}{32.174} * \frac{1}{12^2}\ lbf/in^2

               p = 13.59\  lbft/in^2

P_{atm} = 14.66\  lbf/in^2

so absolute pressure = P_{atm} - p

                                   = 14.66 - 13.59 = 1.07 lbft/in^2

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The mass of a hypothetical planet is 1/100 that of Earth and its radius is 1/4 that of Earth. If a person weighs 600 N on Earth,
omeli [17]

To solve this problem we will apply the Newtonian concept of gravitational acceleration produced by a planet. This relationship is given by:

g = \frac{GM}{r^2}

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The relationship of gravity would then be given:

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The relationship with the new planet, from the gravity of the earth would be given

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The relationship with the weight of the earth would be given as:

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YALL PLS HELP ME! CAN YOU GUYS CHECK MY ANSWERS?
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Answer:

1. Newton's First Law of Motion

2.Newton's Third Law of Motion

3.Newton's Third Law of Motion

4.Newton's Second Law of Motion

5.Newton's Third Law of Motion

6.Newton's Second Law of Motion

7.Newton's First Law of Motion

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Newton's First Law of Motion states that an object remain at rest or moving with a uniform velocity unless an external force acts on it. This is the law of inertia

Newton's Second Law of Motion states that the force of an objects is the product of mass and acceleration.

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4 0
3 years ago
Problem:
kenny6666 [7]

Answer:

a) x = 1.5 *10⁻⁴cos(524πt) m

b) v = -1.5 *10⁻⁴(524π)sin(524πt) m/s

   a =  -1.5 *10⁻⁴(524π)²cos(524πt) m/s²

c) x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

Explanation:

x = Acos(ωt)

ω = 2πf = 2π(262) = 524π rad/s

x = 1.5 *10⁻⁴cos(524πt)

v = y' = -Aωsin(ωt)

v = -1.5 *10⁻⁴(524π)sin(524πt)

a = v' = -Aω²cos(ωt)

a =  -1.5 *10⁻⁴(524π)²cos(524πt)

not sure about the last part as time is generally not given in mm

I will show at 1 second and at 0.001 s to try to cover bases

x(1) = 1.5 *10⁻⁴cos(524π(1))

x(1) = 1.5 *10⁻⁴cos(524π)

x(1) = 1.5 *10⁻⁴(1)

x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = 1.5 *10⁻⁴cos(524π(0.001))

x(0.001) = 1.5 *10⁻⁴cos(0.524π)

x(0.001) = 1.5 *10⁻⁴(-0.0753268)

x(0.001) = -1.129902...*10⁻⁵ m

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

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