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timama [110]
4 years ago
8

An AC circuit with a total resistance of R has an rms current of I. The circuit wraps around an iron core transistor with 50 loo

ps. The secondary coils, leading to a different circuit, wrap with N loops. The secondary circuit has a total resistance of 40 Ω. What is the rms current in the secondary circuit (6 points)?Assigned values for 15.0 Ω R = , I = 5.0 amp, and N = 18
Physics
1 answer:
zvonat [6]4 years ago
6 0

Answer:

current is 13.88 A

Explanation:

given data

no of loops = 50 loops

resistance = 40 Ω

R = 15.0 Ω

I = 5.0 amp

N = 18

to find out

rms current

solution

we will apply here transformed principle that is

v(s) / v(p) = I(p) / I(s) = N(s) / N(p)   ....................1

here p is primary and s is secondary

put here all these value

5 / I(s) = 18 / 50

I(s) = 250 /18

I(s) = 13.88

so current is 13.88 A

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Vector A⃗ points in the positive y direction and has a magnitude of 12 m. Vector B⃗ has a magnitude of 33 m and points in the ne
gavmur [86]

vector A has magnitude 12 m and direction +y

so we can say

\vec A = 12 \hat j

vector B has magnitude 33 m and direction - x

\vec B = -33 \hat i

Now the resultant of vector A and B is given as

\vec A + \vec B = 12 \hat j - 33 \hat i

now for direction of the two vectors resultant will be given as

\theta = tan^{-1}\frac{12}{-33}

\theta = 160 degree

so it is inclined at 160 degree counterclockwise from + x axis

magnitude of A and B will be

R = \sqrt{A^2 + B^2}

R = \sqrt{12^2 + 33^2} = 35.11 m

so magnitude will be 35.11 m

6 0
3 years ago
What are the methods of heat transfer? ​
m_a_m_a [10]

Answer:

Radiation , Conduction and Convection

Explanation:

Those are the ways heat is transferred

8 0
3 years ago
How much work is done by an applied force to lift a 45 newton block 6.0 meters at a constant speed ?
AleksandrR [38]

Answer:

270Joues

Explanation:

Step one:

given data

Force F= 45N

distance moved d= 6m

Required

The work done in moving the block 6m

Step two:

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WD= force* distance

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3 years ago
(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/s² and subway stations are located
andreev551 [17]

Answer:

The correct answer is 32.9 m/s

Explanation:

To solve this, we list out the known and the unknown variables as follows

Maximum allowable acceleration = 1.34 m/s²

Distance between sttions = 806 m

Therefore from the equation of motion

v = ut + 0.5·×at²

Where v = final velocity

u = initial velocity

S = distance covered

t  = time

a = acceleration

Also v² = u² + 2·a·S

where u is the initial velocity, which we can take as u = 0, then

v² = 2·1.34·S = 2.68S m²/s²  then

Also the train has to decelerate from maximum speed to stop at the next tran station wherev = 0, thus v² = u² -2·1.34·Z,  so u² = 2.68Z

since u² = 2.68S from the previous calculation, then for v = 0

2.68S = 2.68Z thus S = Z which and to reach the next subway station S + Z must be = 806 m, then S = 806 m ÷ 2 = 403 m

and v² = 2.68S m²/s² = 1080.04 m²/s²

v = 32.9 m/s

The maximum speed a subway train can attain between stations is 32.9 m/s

3 0
3 years ago
Jennifer and Katie are leaning on each other. Jennifer weighs 150 and Katie weighs 120. Which one is pushing harder on the other
murzikaleks [220]

Jennifer

Explanation:

she has more mass which means she is using more force

3 0
4 years ago
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