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dem82 [27]
3 years ago
15

A solenoid with an inductance of 8 mH is connected in series with a resistance of 5 Ω and an EMF forming a series RL circuit. A

current of 4.0 A is flowing clockwise in the solenoid at time t = 0. The current in the solenoid then grows from 4.0 A to 10.0 A in 0.2 ms. What are the value and the direction of the induced EMF?
Physics
1 answer:
monitta3 years ago
5 0

Answer:

induced EMF = 240 V

and by the lenz's law  direction of induced EMF is opposite to the applied EMF

Explanation:

given data

inductance = 8 mH

resistance = 5 Ω

current = 4.0 A

time t = 0

current grow = 4.0 A to 10.0 A

to find out

value and the direction of the induced EMF

solution

we get here induced EMF of induction is express as

E = - L \frac{dI}{dt}    ...................1

so E = - L \frac{I2 - I1}{dt}

put here value we get

E = - 8 × 10^{-3} \frac{10 - 4}{0.2*10^{-3}}

E = -40 ×  6

E = -240

take magnitude

induced EMF = 240 V

and by the lenz's law we get direction of induced EMF is opposite to the applied EMF

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Montano1993 [528]

Answer:\sqrt{65} m/s

Explanation:

6 0
2 years ago
a rocket of mass 100,000 kg undergoes an acceleration of 2m/s2. show the net force acting on it is 200,000
just olya [345]
M= 100000 kg
a= 2m/s²

Use Newton's second law:
F=ma=100000kg • 2 m/s²
= 200000 N
6 0
3 years ago
2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from
OLEGan [10]

Answer:

Relative\ Velocity = 105m/s

Explanation:

Given

V_A = 45m/s

V_B = 60m/s

Required

Determine the speed of B w.r.t A

The question implies that, we determine the relative velocity of B w.r.t A

Because both trains are moving towards one another, the required velocity is a sum\ of\ velocities\ of both trains:

This is shown below:

Relative\ Velocity = V_A + V_B

Relative\ Velocity = 45m/s + 60m/s

Relative\ Velocity = 105m/s

4 0
3 years ago
Two blocks with mass M and 3M on a horizontal frictionless surface are pushed together and compress a spring of negligible mass
Naily [24]

Answer:

launching speed of the lighter block = -6 m/s

Explanation:

We are given;

Mass of light block; M

Mass of heavy block; 3M

Speed of launched block: v = 2m/s

We are told that the two blocks are sitting on the horizontal frictionless surface. Thus, we can say that no external force is being applied on the system and so, the momentum of the whole system is conserved accordingly to that condition.

We are also told that when the rope is cut with scissors, that the heavier block attains the speed of 2 m/s in the positive x-direction which is horizontal direction.

We know that formula for momentum is; M = mass x velocity.

Thus, the momentum of the heavier block is calculated as;

M_1 = 3M × 2

M_1 = 6M kg.m/s

Since no external force is applied on the object, the initial momentum will be zero.

Hence, to conserve the system, the momentum of the lighter block will be equal and opposite to the momentum of heavier block.

So, momentum of lighter block is;

M_2 = -6M kg.m/s

Since mass of lighter block is M and formula for momentum = mass x velocity.

Thus;

-6M = Mv

Where v is speed of lighter block.

So, v = -6M/M

v = -6 m/s

7 0
3 years ago
A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.
Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

v^{2}=u^{2}+2as

0^{2}=35^{2}- 2 \times 9.8 \times h

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

By use of first equation of motion

v = u + at

0 = 35 - 9.8 t

t = 3.57 s

The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

8 0
3 years ago
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