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Nataly_w [17]
3 years ago
15

A square coil of wire with side 8.0 cm and 50 turns sits in a uniform magnetic field that is perpendicular to the plane of the c

oil. The coil is pulled quickly out of the magnetic field in 0.2 s. If the resistance of the coil is 15 ohm and a current of 12 mA is induced in the coil, calculate the value of the magnetic field.
Physics
1 answer:
bezimeni [28]3 years ago
8 0

Answer:

Explanation:

area of the coil  A = .08 x .08 = 64 x 10⁻⁴ m ²

flux through the coil Φ = area of coil x no of turns x magnetic field

= 64 x 10⁻⁴ x 50 x B where B is magnetic field

emf induced = dΦ / dt = ( 64 x 10⁻⁴ x 50 x B - 0 ) / .2

= 1.6 B

current induced = emf induced / resistance

12 x 10⁻³ = 1.6 B / 15

B = 112.5 x 10⁻³ T .

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Answer:

m1/6 ( c )

Explanation:

since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy

After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3   ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3  ----- ( 3 )

All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3

note ; 3p = m1v ∴ m3 = \frac{m1v}{3v3}  -----  ( 4 )

applying conservation of energy

3v = v1 + v2 + v3 ------- ( 5 )

also 3m1v1 = m1v = v1 = v/3 =

v2 + v3 = 8/3 v ----- ( 6 )

next eliminate V3 for equation 6 by applying conservation of energy and momentum

m1 =  2m2 ------ ( 7 )

now using p1 = p2 = m1v1 = 1/2 m1v1  hence v2 = 2v1  where v1 = 1/3 v

hence ; v2 = 2/3 v ------- ( 8 )

solving with equation 6 and 8

v3 = 2v ------ ( 9 ) ∴  v/v3 = 1/2 ---- ( 10 )

solving with equation 9 and 10

m3 = m1/3 * 1/2 = m1/6

8 0
3 years ago
A dragster in a race accelerated from rest to 60 m/s by the time it reached the finish line. The dragster moved the distance fro
Andreyy89

Answer:

l don't now but l think the is 160

Explanation:

160 or 810

6 0
3 years ago
In order for an external force to do work on a system, Question 9 options: a) there must be a component of the force perpendicul
zepelin [54]

Answer:

b) there must be a component of force parallel to the motion of the object.

Explanation:

We know that work done on a body by an external force is calculated by the formula given below:

W = F.d = Fd Cos θ

where,

W = Work Done by the force on the body

F = Magnitude of force

d = displacement of the body

θ = The angle between the direction of motion of the body and the force applied

It is clear from the formula of the work done, that "F Cosθ" represents the component of the force, that is acting in the direction of motion of the object or parallel to the direction of motion of the object. So, if there is no component of force parallel to motion of object, then this factor will become zero. As a result, the work done will also be zero.

Therefore, the correct option will be:

b) <u>there must be a component of force parallel to the motion of object.</u>

6 0
3 years ago
A point charge Q at the center of a sphere of radius R produces an electric flux of Φ coming out of the sphere. If the charge r
Dafna1 [17]

Remains the same

Explanation:

According to Gauss's law, the electric flux through a closed surface is proportional to the charge enclosed by the surface. So no matter how big or small we make the surface that encloses the charge, the electric flux remains the same because it only depends on the enclosed charge, not surface area.

3 0
2 years ago
Calculate the electric field strength required to just support an oil drop of mass
Ray Of Light [21]

Answer:

E= 3.06 \times 10^{-17} N/C

Explanation:

As we know that the oil drop experiment tells us the balancing the gravitational force by electrostatic force.

So ,

Electrostatic force = Gravitational force

qE=mg

Here , q is charge , E is electric field , m is mass and g is gravity.

so,

E=\frac{mg}{q}

Insert values from question,

E=\frac{10^{-2}kg \times 9.8 m/s^{2}}{3.2 \times 10^{-19}C}

or

E= 3.06 \times 10^{-17} N/C

This is the required value of electric field.

3 0
3 years ago
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