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3 years ago

7

a 4m long straight wir that carries acurrent of 0.5A is placed perpendicular to a uniform magnetic field. if the size of magneti

c deflection force is 0.4N, then what is the strength of magbetic field?

1 answer:

PolarNik [594]3 years ago

5 0

Answer:

B=0.2T

Explanation:

given required solution

l=4m B=? <em>F</em><em>=</em><em>BIL</em>

i=0.5A B=F/IL

F=0.4N B=0.4N/0.5A*4m

B=0.4/2=0.2T

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While traveling on a horizontal road at speed vi, a driver sees a large rabbit ahead and slams on the brakes. The wheels lock an

MA_775_DIABLO [31]

Answer:

μk = (Vf - Vc)/(T×g)

Explanation:

Given

Vi = initial velocity of the car

Vf = final velocity of the car

T = Time of application of brakes

g = acceleration due to gravity (known constant)

Let the mass of the car be Mc

Assuming only kinetic frictional force acts on the car as the driver applies the brakes,

The n from Newtown's second law of motion.

Fk = Mc×a

Fk = μk×Mc×g

a = (Vf - Vc)/T

Equating both preceding equation.

μk×Mc×g = Mc × (Vf - Vc)/T

Mc cancels out.

μk = (Vf - Vc)/(T×g)

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3 years ago

The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

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3 years ago

A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wave

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Answer:

$\lambda$ = 40 cm

Explanation:

given data

string length = 30 cm

solution

we take here equation of length that is

L = $n \times \frac{1}{4} \lambda$ ...............1

so

total length will be here

$L = \frac{\lambda}{2} + \frac{\lambda}{4}\\$

$L = \frac{3 \lambda }{4}$

so $\lambda$ will be

$\lambda = \frac{4L}{3}\\\lambda = \frac{4\times 30}{3}$

$\lambda$ = 40 cm

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3 years ago

A ball has a mass of 2g and a velocity of 3m/s. What is the ball's Kinetic Energy?

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Answer: im not sire

Explanation: very sorry im not sure

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pav-90 [236]

if in series one lightbulb burns out the rest are unable to turn on.

In parallel a single light bulb burns out any other light bulbs are able to work.

Parallel is the best to use during holidays.

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3 years ago