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ira [324]
3 years ago
9

Find the capacitance reactance of a 0.1 micro frequency capacitor 50Hz and at 200Hz​

Engineering
1 answer:
Paraphin [41]3 years ago
5 0

Complete Question:

Find the capacitive reactance of a 0.1 microfarad capacitor with frequency 50Hz and at 200Hz​.

Answer:

I. Capacitive reactance = 31826.86 Ohms.

II. Capacitive reactance = 7956.72 Ohms.

Explanation:

<u>Given the following data;</u>

Capacitance = 0.1 uF = 0.0000001 Farad

Frequency = 50 Hz and 200 Hz

To find the capacitive reactance;

Mathematically, the capacitive reactance of an electronic circuit is given by the formula;

X_{c} = \frac {1}{2 \pi fc}

Where;

  • Xc is capacitive reactance.
  • f is the frequency.
  • c is the capacitance.

Substituting into the formula, we have;

I. At frequency, f = 50 Hz

X_{c} = \frac {1}{2 * 3.142 * 50 * 0.0000001}

X_{c} = \frac {1}{314.2 * 0.0000001}

X_{c} = \frac {1}{0.00003142}

X_{c} = 31826.86

II. At frequency, f = 200 Hz

X_{c} = \frac {1}{2 * 3.142 * 200 * 0.0000001}

X_{c} = \frac {1}{1256.8 * 0.0000001}

X_{c} = \frac {1}{0.00012568}

X_{c} = 7956.72

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If a torque of M = 300 N⋅m is applied to the flywheel, determine the force that must be developed in the hydraulic cylinder CD t
Licemer1 [7]

Answer:

<em>866.1 N</em>

Explanation:

The torque on the flywheel = 300 N-m

The force from the hydraulic cylinder will generate a moment on CA about point A.

The part of this moment that will be at point B about A must be proportional to the torque on the cylinder which is 300 N-m

we know that moment = F x d

where F is the force, and

d is the perpendicular distance from the turning point = 1 m

Equating, we have

300 = F x 1

F = 300 N   this is the frictional force that stops the flywheel

From F = μN

where F is the frictional force

μ is the coefficient of static friction = 0.4

N is the normal force from the hydraulic cylinder

substituting, we have

300 = 0.4 x N

N = 300/0.4 = 750 N

This normal force calculated is perpendicular to CA. This actual force, is at 30° from the horizontal. To get the force from the hydraulic cylinder R, we use the relationship

N = R sin (90 - 30)

750 = R sin 60°

750 = 0.866R

R = 750/0.866 = <em>866.1 N</em>

3 0
3 years ago
A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
kompoz [17]

Answer:

a. The mass flow rate (in lbm/s) is 135lbm/s

b. The temperature (in o F) is 200.8°F

Explanation:

We assume that potential energy and kinetic energy are negligible and the control volume operates at a steady state.

Given

a. The mass flow rate (in lbm/s) is 135lbm/s

b.

m1 = Rate at inlet 1 = 125lbm/s

m2 = Rate at inlet 2 = 10lbm/s

The mass flow rate (in lbm/s) is calculated as m1 + m2

Mass flow rate = 125lbm/s + 10lbm/s

Mass flow rate = 135lbm/s

Hence, the mass flow rate (in lbm/s) is 135lbm/s

b. To calculate the temperature.

First we need to determine the enthalpy h1 at 14.7psia

Using table A-3E (thermodynamics)

h1 = 180.15 Btu/Ibm

h2 at 14.7psia and 60°F = 28.08 Btu/Ibm

Calculating h3 using the following formula

h3 = (h1m1 + h2m2) / M3

h3 = (180.15 * 125 + 28.08 * 10)/135

h3 = 168.8855555555555

h3 = 168.89 Btu/Ibm

To get the final temperature; we make use of table A-2E of thermodynamics.

Because h3 < h1, it means the liquid is at a compressed state.

The corresponding temperature at h3 = 168.89 is 200.8°F

The temperature (in o F) is 200.8°F

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3 years ago
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3 years ago
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Ordan has _ 5 8 can of green paint and _ 3 6 can of blue paint. If the cans are the same size, does Jordan have more green paint
Morgarella [4.7K]

Answer:

Jordan has more green paints

Explanation:

Given

Green = \frac{5}{8}

Blue = \frac{3}{6}

Required

Which paint does he have more?

For better understanding, it's better to convert both measurements to decimal.

For the green paint:

Green = \frac{5}{8}

Green = 0.625

For the blue paint:

Blue = \frac{3}{6}

Blue = 0.5

By comparison:

0.625 > 0.5

<em>This means that Jordan has more green paints</em>

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A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of he
hichkok12 [17]

Answer:

ΔT=  11.94 °C

Explanation:

Given that

mass of water = 10 kh

Time t= 15 min

Heat lot from water = 400  KJ

Heat input to the water = 1  KW

Heat input the water= 1 x 15 x 60

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By heat balancing

Heat supply - heat rejected = Heat gain by water

As we know that heat capacity of water

C_p=4.187 \frac{KJ}{kg-K}

Q=mC_p\Delta T

Now by putting the values

900 - 400 = 10 x 4.187 x ΔT

So  rise in temperature of water ΔT=  11.94 °C

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