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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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Two trumpet players are trying to tune their instruments. When they are in tune, they will both be playing the same note and no
marusya05 [52]

Answer:

The second trumpeter will be playing at frequency = 515 Hz

Explanation: Given that the note sounds lower and they can hear 20 beats in 4.0 s. 

Beat frequency = 20/4 = 5 Hz

Beat frequency = F2 - F1

5 = 520 - F1

F1 = 520 - 5

F1 = 515 Hz

Since the note sound lower, the second trumpeter will be playing at 515 Hz frequency

8 0
3 years ago
At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/
Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

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One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

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Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

4 0
3 years ago
When light of wavelength 240 nm falls on a potassium surface, electrons having a maximum kinetic energy of 2.93 eV are emitted.
olchik [2.2K]

A) We want to find the work function of the potassium. Apply this equation:

E = 1243/λ - Φ

E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function

Given values:

E = 2.93eV, λ = 240nm

Plug in and solve for Φ:

2.93 = 1243/240 - Φ

Φ = 2.25eV

B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:

E = 1243/λ - Φ

0 = 1243/λ - Φ

0 = 1243/λ - 2.25

λ = 552nm

C) We want to find the frequency associated with the threshold wavelength. Apply this equation:

c = fλ

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Given values:

c = 3×10⁸m/s, λ = 5.52×10⁻⁷m

Plug in and solve for f:

3×10⁸ = f(5.52×10⁻⁷)

f = 5.43×10¹⁴Hz

7 0
3 years ago
Which temperature scale has the highest value for the boiling point of water?
drek231 [11]

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The expression for the conversion of degree Celsius to Kelvin is as follows:

K= 273 + degree Celsius

The expression for the conversion of degree Celsius to Fahrenheit is as follows:

F=(Degree Celsius\times \frac{9}{5})+ 32

The freezing point of water on Celsius degree is zero degree Celsius. The freezing point on kelvin scale is 273.15 K.

The boiling point of water on Celsius degree is 100 degree Celsius. The boiling point on kelvin scale is 373.15 K.  The boiling point on Fahrenheit scale is 373.15 K.

Therefore, Kelvin scale has the highest value for the boiling point of water.

6 0
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Solar winds have an influence all the way to about 160 AU from the Sun. About how many kilometers is that?
Mumz [18]

Answer: A. 24,000,000,000 km

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Astronomical unit is mainly used to measure distances within our solar system.

1 AU is equal to  150\times 10^6 km

Thus 160 AU  is equal to \frac{150\times 10^6}{1}\times 160=24000\times 10^6km=24000000000km

4 0
2 years ago
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