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3 years ago

Please Help, I need to turn this in today

Physics

2 answers:

Scrat [10]3 years ago

4 0

You're right, its C. Good work

mart [117]3 years ago

3 0

I think it is C , idk why, but I just learn about this and I think it is.... sorry I didn't really help.....

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Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 430 and 610 turns, respectively. A

KIM [24]

Answer

given,

Two solenoids A and B

Number of turn

Na = 430 turns Nb = 610 turns

Current = 2.80 A

Average flux through A = 300 μWb

Average of flux through B = 90.0 μ Wb

a) $L = \dfrac{N \phi}{I}$

$L = \dfrac{610\times 90 \times 10^{-6}}{2.80}$

$L =19.6 mH$

b) inductance of A

$L = \dfrac{N_A \phi_A}{I_A}$

$L = \dfrac{430\times 300 \times 10^{-6}}{2.80}$

$L =46 mH$

c) magnitude of the emf

$\epsilon_B = -L_B\dfrac{dI}{dT}$

$\epsilon_B = -(19.6\times 10^{-3})(0.5)$

$\epsilon_B = -9.8\times 10^{-3}\ V$

$\epsilon_B = -9.8\ mV$

7 0

3 years ago

If an object is rolling without slipping, how does its linear speed compare to its rotational speed? If an object is rolling wit

salantis [7]

Answer:

$v = r\omega$

Explanation:

If the object is rolling without slipping, every unit of rotated angle equals to a distance perimeter rotated.

Suppose the object complete 1 revolution within time t. The angular distance is 2π rad. Its angular velocity is 2π/t

The distance it covered is its circumference, which is 2πr, and so the speed is 2πr/t

So the linear speed compared to angular speed is

$\frac{v}{\omega} = \frac{2\pi r/t}{2\pi /t} = r$

$v = r\omega$

6 0

3 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

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2 years ago

If earth's axis was tilted 45 degrees at what latitudes

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23.5 DEGREES is the answer.

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2 years ago

A speaker draws 2 amps of current when it is connected to a 16 volt source. What is the resistance of the speaker? Question 2 op

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