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kirza4 [7]
3 years ago
12

A coil has an inductance of 6.00 mH, and the current in it changes from 0.200 A to 1.50 A in a time interval of 0.300 s. Find th

e magnitude of the average induced emf in the coil during this time interval.
Physics
1 answer:
Andre45 [30]3 years ago
5 0

Answer:

0.026 V

Explanation:

Given that,

Inductance of the coil, L = 6 mH

The current changes from 0.2 A to 1.5 A in a time interval of 0.3 s

We need to find the magnitude of the average induced emf in the coil during this time interval. The formula for the induced emf is given by :

\epsilon=L\dfrac{dI}{dt}\\\\=6\times 10^{-3}\times \dfrac{1.5-0.2}{0.3}\\\\=0.026\ V

So, the magnitude of induced emf is 0.026 volts.

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What is its speed after 3. 83 as if it accelerates uniformly at −3. 04 m/s 2 ? answer in units of m/s.
weeeeeb [17]

The velocity equation is v_{final} =v_{initial} +at\\

Known facts:

  • t = 3.83s
  • a= -3.04
  • intial velocity = 0

Plug into equation known quantities:

   v_{final} = (-3.04) * 3.83 = -11.6432m/s

Thus the final velocity is -11.6432m/s

Hope that helps!

6 0
3 years ago
two lenses are combined together . if the power of one of the lenses is +5D and combined focal length is 0.4m.What is the power
s344n2d4d5 [400]

Answer:

-0.4D(maybe)

Explanation:

combined focal length (f)= 0.4m

D1 = 5D

Then f1= 1/D1

= 1/5 = 0.2 m

1/f=1/f1+1/f2

1/0.4=1/0.2+1/f2

f2= -5/2

D2=1/f2= -0.4D

8 0
2 years ago
Which cars have the same magnitude of momentum? Check all that apply. car 1 car 2 car 3 car 4 car 5
wariber [46]

Answer:

Car 1 and Car 2 have the same momentum!

Explanation:

Using the formula of momentum (P=m*v), we get for each car:

Car 1: 5kg*2.2m/s = 11kg*m/s

Car 2: 5.5kg*2m/s = 11kg*m/s

Car 3: 6kg*1.35m/s = 8.1kg*m/s

Car 4: 6.5kg*1.9m/s = 12.35kg*m/s

Car 5: 7kg*1.25m/s = 8.75kg*m/s

7 0
3 years ago
What is latent heat energy's definiton
tamaranim1 [39]
The heat energy which has to be supplied to change the state of a substance is called its latent heat. Latent heat does not increase the kinetic energy of the particles of the substance, so the temperature of a substance does not rise during the change of state. :))
6 0
3 years ago
A catapult with a radial arm 3.81 m long accelerates a ball of mass 18.2 kg through a quarter circle. The ball leaves the appara
saw5 [17]

Answer:

(a)\alpha = 53.73 m/s^2

(b)   I =428 kgm^2

(c)\tau = 428 \times 53.73  = 22996 .44Nm

Explanation:

GIVEN

mass = 18.2 kg

radial arm length = 3.81 m

velocity = 49.8 m/s

mass of arm = 22.6 kg

we know using relation between linear velocity and angular velocity

\omega = \frac{v}{l}

\omega = \frac{49.8}{3.81} \\\omega = 12.99 rad/s

for  angular acceleration, use the following equation.

\omega _{f}^2 = \omega_{i}^2+2\alpha\theta

since \omega _{i} = 0

here  for one circle is 2 π radians.   therefore for one quarter of a circle is π/2 radians

so   for one quarter \theta = \frac{\pi }{2}

(12.99)^2 = 2\alpha(\frac{\pi }{2})

on solving

\alpha = \frac{168.74}{\pi }\\\alpha = 53.73 m/s^2

(b)

For the catapult,

moment of inertia

I = \frac{1}{2}MR^2

I = \frac{1}{2} \times 22.6\times 3.81 \times 3.81\I = 164kg m^2

For the ball,

I = MR^2

I = 18.2 \times 14.51

I = 264 kgm^2

so total moment of inertia =  428 kgm^2

(c)

\tau = I\alpha

\tau = 428 \times 53.73  = 22996 .44Nm

3 0
4 years ago
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