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Blizzard [7]
3 years ago
10

a 100mh inductor is placed parallel with a 100 ohm resistor in the circuit , the circuit has a source voltage of 30 vac and a fr

equancy of 200 Hz what is the current through the inductor
Engineering
1 answer:
ddd [48]3 years ago
5 0

Answer:

current through the inductor  = 0.24 A

Explanation:

given data

inductor = 100mh

resistor = 100 ohm

voltage = 30 vac

frequancy = 200 Hz

to find out

current through the inductor

solution

current through the inductor will be when inductor is place parallel with resistor

across resistor

XL = i2πl

XL = i2π×200×100×10^{-3} = i126.66

so

current through the inductor = \frac{voltage}{XL}

current through the inductor = \frac{30}{125.66}

current through the inductor  = 0.24 A

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A system has a level 1 cache and a level 2 cache. The hit rate of the level 1 cache is 90% and the hit rate of the level 2 cache
Hatshy [7]

Answer:

B) 2.22

Explanation:

In a first step, the system will look in cache number 1. If it is not found in cache number 1, then the system will look in cache number 2. Finally (if not in cache 2 also) the system will look in main memory.

The average access time will take into consideration success in cache 1, failure in cache number 1 but success in cache number 2, failure in cache number 1 and 2 but success in main memory.

Mathematically we can write it into the following equation :

AAT=(H1.T1)+(1-H1).H2.T2+(1-H1).(1-H2).Hm.Tm

Where AAT is the average access time

H1,H2 and Hm are the hit rate of cache 1,cache 2 and main memory respectively.

T1,T2 and Tm are the access time of cache 1, cache 2 and main memory respectively

Hm = 1

AAT=(0.9).(1)+(1-0.9).(0.8).(4)+(1-0.9).(1-0.8).(1).(50)=0.9+0.32+1=2.22

AAT=2.22

Therefore, option b) is the correct.

6 0
3 years ago
Beryllium (Be) has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.568. If the radius of the Be atom is
sweet-ann [11.9K]

Answer:

Unit cell volume will be

4.866*10^{-2}nm^{3}

8 0
3 years ago
An inventor claims to have developed a refrigerator that at steady state requires a net power input of 1.1 horsepower to remove
Lynna [10]

Answer:

The inventor's claim is false in the sense that no thermal machine can violate the first thermodynamic law.

Explanation:

The inventor's claim could not be possible as no thermal machine can transfer more heat than the input work consumed. If we expose the thermal efficiency:

n=Q out / W in

Where Q and W both must be in the same power unit, so we will convert the remove heat from BTU/hr to hp:

12000 BTU/hr = 4.72 hp

Therefore by comparing, we notice that the removing heat of 4.75 hp is large than the delivered work of 1.11 hp. By evaluating the efficiency:

[tex]n=4.75 hp / 1.1 hp  = 4.3 > 1[/tex]

6 0
3 years ago
An object of mass 521 kg, initially having a velocity of 90 m/s, decelerates to a final velocity of 14 m/s. What is the change i
Harman [31]

Answer:2058.992KJ

Explanation:

Given data

Mass of object\left ( m\right )=521kg

initial velocity\left ( v_0\right )=90m/s

Final velocity\left ( v\right )=14m/s

kinetic energy of body is given by=\frac{1}{2}mv^{2}

change in kinectic energy is given by substracting  final kinetic energy from initial kinetic energy of body.

Change in kinetic energy=\frac{1}{2}\times m\left ( V_0^{2}-V^2\right )

Change in kinetic energy=\frac{1}{2}\times521\left ( 90^{2}-14^2\right )

Change in kinetic energy=2058.992KJ

7 0
3 years ago
The high-pressure air system at OSU's Aerospace Research Center is fed by a set of two cylindrical tanks. Each tank has an outer
lana66690 [7]

Answer:

179000 lb

Explanation:

The supports must be able to hold the weight of the tank and the contents. Since tanks are pressure tested with water, and the supports cannot fail during testing, we disregard the air and will consider the weight of water.

The specific weight of water is ρw = 62.4 lbf/ft^3

These tanks are thin walled because

D / t = 4.6 / 0.1 = 46 > 10

To calculate the volume of steel we can approximate it by multiplying the total surface area by the thickness:

A = 2 * π/4 * D^2 + π * D * h

The steel volume is:

V = A * t

The specific weight is

ρ = δ * g

ρs = 499 lbm/ft^3 * 1 lbf/lbm = 499 lbf/ft^3

The weight of the steel tank is:

Ws = ρs * V

Ws = ρs * A * t

Ws = ρs * (2 * π/4 * D^2 + π * D * h) * t

Ws = 499 * (π/2 * 4.6^2 + π * 4.6 * 50) * 0.1 = 37700 lb

The weight of water can be approximated with the volume of the tank:

Vw = π/4 * D^2 * h

Ww = ρw * π/4 * D^2 * h

Ww = 62.4 * π/4 * 4.6^2 * 50 = 51800 lb

Wt = Ws + Ww = 37700 + 51800 = 89500 lb

Assuming the support holds both tanks

2 * 89500 = 179000 lb

The support must be able to carry 179000 lb

3 0
4 years ago
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