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tester [92]
3 years ago
6

Use cylindrical coordinates. Evaluate E (x − y) dV, where E is the solid that lies between the cylinders x2 + y2 = 1 and x2 + y2

= 16, above the xy-plane, and below the plane z = y + 4. g
Physics
1 answer:
S_A_V [24]3 years ago
3 0

Answer:hchv nhgjj

Explanation:

ggchh

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An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made fro
Dmitrij [34]

Answer:

(a) 17634.24 Ω

(b) 0.0068 A

Explanation:

(a)

The formula for inductive inductance is given as

X' = 2πFL................... Equation 1

Where X' = inductive reactance, F = frequency, L = inductance

Given: F = 60 Hz, L = 46.8 H, π = 3.14

Substitute into equation 1

X' = 2(3.14)(60)(46.8)

X' = 17634.24 Ω

(b)

From Ohm's law,

Vrms = X'Irms

Where Vrms = Rms Voltage, Irms = rms Current.

make Irms the subject of the equation

Irms = Vrms/X'...................... Equation 2

Given: Vrms = 120 V, X' = 17634.24 Ω

Substitute into equation 2

Irms = 120/17634.24

Irms = 0.0068 A

5 0
3 years ago
Which of the following is the only requirement for an object to be in projectile motion? A. The horizontal and vertical motions
NARA [144]
The answer is d, gravity is the only force acting on the object
8 0
3 years ago
What is the unit of velocity ratio and mechanical advantage and why​
alexandr402 [8]
Both don’t have units beacuse they are ratios
8 0
3 years ago
Why are the freezing points of water and the melting point the same
Sladkaya [172]
Because melting point<span> and </span>freezing point<span> describe the</span>same<span> transition of matter, in this case from liquid to solid (</span>freezing) or equivalently, from solid to liquid (melting<span>).</span>
5 0
3 years ago
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An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b
Mariana [72]

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

6 0
3 years ago
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