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nasty-shy [4]
3 years ago
11

Coils of various dimensions designed to introduce specified amounts of inductance into a circuit are called (a) semiconductors (

b) superconductors (c) electromagnets (d) inductons
Engineering
1 answer:
Delicious77 [7]3 years ago
4 0

Answer: d) Inductors

Explanation:  Inductors are the electric component commonly known by the name of coils as well. The work done by inductors is when the electric current flows through the coils of the wire then there is production of the magnetic field. They are the component that have the inductance in particular amount according to the circuit.So, coils of various dimensions are designed to introduce specified amount of inductance  into a circuit are called inductors.

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Which lens is wide-angle?
PIT_PIT [208]

A wide-angle lens has a focal length of 35mm or shorter, which gives you a wide field of view. The wider your field of view, the more of the scene you'll be able to see in the frame. These lenses are ideal for many scenarios, and most photographers have at least one trusty wide-angle lens in their kit.

BRAINLIEST PLSSS

5 0
2 years ago
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
3 years ago
three balls each have a mass m if a has a speed v just before a direct collision with B determine the speed of C after collision
ratelena [41]

Answer:

Vc2= V(l+e) ^2/4

Vg2= V(l-e^2)/4

Explanation:

Conservation momentum, when ball A strikes Ball B

Where,

M= Mass

V= Velocity

Ma(VA)1+ Mg(Vg)2= Ma(Va)2+ Ma(Vg)2

MV + 0= MVg2

Coefficient of restitution =

e= (Vg)2- (Va)2/(Va)1- (Vg)1

e= (Vg)2- (Va)2/ V-0

Solving equation 1 and 2 yield

(Va)2= V(l-e) /2

(Vg)2= V(l+e)/2

Conservative momentum when ball b strikes c

Mg(Vg)2+Mc(Vc)1 = Mg(Vg)3+Mc(Vc)2

=> M[V(l+e) /2] + 0 = M(Vg)3 + M(Vc) 2

Coefficient of Restitution,

e= (Vc)2 - (Vg)2/(Vg)2- (Vc)1

=> e= (Vc)2 - (Vg)2/V(l+e) /2

Solving equation 3 and 4,

Vc2= V(l+e) ^2/4

Vg2= V(l-e^2)/4

8 0
3 years ago
Read 2 more answers
Which of the following statements are true concerning AC circuit that contains both resistance and inductance? A. The current an
maw [93]

The current will lag the voltage in AC circuit that contains both resistance and inductance.

Answer: C

Explanation

There is no inductance only circuits in reality.

The circuits containing inductance has also a lower amount of resistance.

The current flows in both resistance and inductance.

There is a drop in the total voltage in resistance and inductance giving rise to the voltage applied in the coil when connected in a series.

An example being inductance coil an AC circuit connected to both resistance and inductance in series.

From the vector diagram, this conclusion can be drawn.

3 0
3 years ago
what is the transfer function of the loaded filter? express your answer in terms of the variables r , l , rl , and s .
NISA [10]

Loaded, H_{Loaded}(s) = \frac{RR_{L} }{R+R_{L} } /(\frac{RR_{L} }{R+R_{L} }+SL) = \frac{RR_{L}/L }{R+R_{L} } /(\frac{RR_{L} /L}{R+R_{L} }+S) is the loaded filter's transfer function.

A graded filter that, by virtue of its weight and permeability, stabilises the foot of an earth dam or other construction when it is installed at the base of that structure.

Air filters with depth loaded are made to achieve precisely that. They add particles gradually to create air passageways, reducing constriction. You may save time and money by using filters that last longer thanks to them. The bigger particles are caught at the filter's beginning, while the smaller particles are caught as it gets closer. This is intended to avoid rapid surface loading, hence facilitating more airflow. This enables longer-lasting filtration as well.

On the other hand, surface loading filters catch every particle that is on its surface. No matter how big or little the particles are, it doesn't care.

Learn more about Loaded here:

brainly.com/question/20039214

#SPJ4

3 0
1 year ago
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