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amid [387]
4 years ago
9

How do you change current in a circuit without changing the voltage?

Physics
1 answer:
Sunny_sXe [5.5K]4 years ago
4 0

Decrease the reactance is the correct answer i believe

Explanation:

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In these chemical equations, each of the letters J through M represents a different element. Which equation is balanced?
Kobotan [32]

Answer:

D. 2JK3 + 3L2M --> 6LK + J2M3

Explanation:

5 0
3 years ago
What is meant by the term blue moon
Andreyy89
“Once in a blue moon” is a common expression that has been used for a long time, and which means 'not very often,' or 'very rarely. ' It often refers to an extra full moon; however, it has been used to describe the way the moon actually looked, when for different reasons it had turned a blueish color.
www.loc.gov › astronomy › item
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4 0
3 years ago
A heat engine operates with 70.0 kcal of heat supplied and exhausts 30.0 kcal of heat. How much work did the engine do?
gladu [14]

The work done by the heat engine is 40 kCal.

The given parameters;

  • input heat of the engine, Q₁ = 70 kCal
  • output heat of the engine, Q₂ = 30 kCal

To find:

  • the work done by the heat engine

The work done by the heat engine is the change in the heat energy of the engine;

W = Q₂ - Q₁

Substitute the given parameters and solve work done (W)

W = 70 kCal - 30 kal

W = 40 kCal

Thus, the work done by the heat engine is 40 kCal.

Learn more here: brainly.com/question/4280097

4 0
3 years ago
Two bicycle tires are set rolling with the same initial speed of 4.00 m/s along a long, straight road, and the distance each tra
77julia77 [94]

Answer:

The coefficient of rolling friction for the tire under low pressure is 0.0342.

Explanation:

Two bicycle tires are set rolling with the same initial speed of 4.00 m/s

Final speed of both the bicycle, speed is reduced by half is measured, v = 2 m/s.

Here,

u_kmg = ma\\\\a=\mu g

Using third equation of motion as :

v^2-u^2=2as\\\\v^2-u^2=2\mu gs\\\\\mu =\dfrac{v^2-u^2}{2gs}\\\\\mu =\dfrac{4^2-2^2}{2\times 9.8\times 17.9}\\\\\mu=0.0342

So, the coefficient of rolling friction for the tire under low pressure is 0.0342.

5 0
3 years ago
A circular loop of radius 13 cm carries a current of 16 A. A flat coil of radius 0.63 cm, having 48 turns and a current of 1.5 A
azamat

Answer:

a) Bt = 7.73 * 10^-5 T

b) T = 6.94 * 10^-7 N*m

Explanation:

Step 1: Data given

Circumar loop Radius = 13 cm

Current = 16 A

Flat coil radius = 0.63 cm

48 turns

Current = 1.5 A

<em> a) What is the magnitude of (a) the magnetic field produced by the loop at its center</em>

Let's assume a loop concentric with a coil, the plane of the coil is perpendicular to the plane of the loop. The magnetic field due to the loop at the center of the loop can be given by:

Bt = µ0It / 2Rt

In this case we'll get:

Bt = ((4π * 10^-7 T*m/A)(16A)) /(2*0.13m)

<u>Bt = 7.73 * 10^-5 T</u>

<em> b) What is the magnitude of the torque on the coil due to the loop?</em>

The torque magnitude excreting on the coil due to the magnetic field of the loop is given by:

T = µcBtsin(∅)

with µc = the magnetic dipole moment of the coil

with ∅ = the angle between the magnetic dipole moment and the magnetic field. The magnetic dipole moment is given by:

µc = N*Ic*A

⇒ with N = the number of turns in the coil

⇒ with A =  πRc² = the area of the coil

µc =π*N*Ic*Rc²

T= π*N*Ic*Rc²*Bt(sin∅)

In this situation we'll have:

T= π*48*1.5A* (0.63 *10^-2m)²*(7.73 * 10^-5 T)*sin(90)

T = <u>6.94 * 10^-7 N*m</u>

8 0
3 years ago
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