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spin [16.1K]
3 years ago
15

Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surfac

e is tilted at an angle of from the horizontal, and a net magnetic flux of 8.4 103 Wb passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. (b) Find the smallest possible value for the area of the second surface.
Physics
1 answer:
alexgriva [62]3 years ago
7 0

Answer:

a. A = 0.0859 m^2

b. A = 0.0178 m^2

Explanation:

Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surface is tilted at an angle of from the horizontal, and a net magnetic flux of 8.4 103 Wb passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. (b) Find the smallest possible value for the area of the second surface.

take note that the question has not specified th angle which the surface is tilted so i assume the angle is at 12^{0} to the horizontal

flux = BAcos(\alpha)

B=magnetic flux in Weber

A=area of the flat surface in m^2

\alpha=the angle to the horizontal

a) 8.4 x10^-3= (.47)Acos(78)

alpha has to be the angle from the normal and not the horizontal so 90-12=78,

 8.4 x10^-3

/(.47)cos(78)

A = 0.0859 m^2

b) If flux remains the same then for it to be the smallest possible area it needs to be perpendicular to the magnetic field so alpha would be 0.

8.4 x10^-3 = (.47)Acos(0)

A = 0.0178 m^2

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Explanation:

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BRAINLIEST AND 30 POINTS‼️‼️
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802.3kilo joules of energy represents 191.75 kilo calorie

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2 years ago
A force of 1200 N is applied to a drum of radius 80 cm which has a 12 kg mass, m1, attached by a cord
maksim [4K]

Answer:

1.022 x 103 N.m

Explanation:

Solution

Given:

The weight of the block of mass m₂ is :

w₂ = m₂*g

Where

w₂ = 39 x 9.8 = 382.2 N

Then,

The weight of the block of mass m₁

w₁= m₁*g;

so,

w₁ = 12 x 9.8 = 117.5 N

Thus,

The tension wrapped in cord on drum (80 cm) T₁ = F - w₁

Now,

T₁ = 1200 - 117.5

T₁ = 1082.5 N

The tension  wrapped in the cord on drum (41 cm) T₂ = w₂;

T₂ = 382.2 N

Hence,

We calculate net torque on the center of the drum:

The net torque = T₁ x 0.8 + T₂ x 0.41;

= 1082.5 x 0.8 + 382.2 x 0.41;

= 1.022 x 103 N.m

Therefore, the resulting torque applied to the system is 1.022 x 103 N.m

4 0
3 years ago
Dog starts at position X equals 2.50 m and undergoes displacement of 8.25 m what is its final position
BabaBlast [244]

Answer:

displacement is how far the dog is from the original position so use addition

Explanation:

2.50+8.25= 10.75

4 0
2 years ago
a current of 1.80 a flows in a wire. how many electrons are flowing past any point in the wire per second?
liubo4ka [24]

Any point in the wire has 1.12 x 10^{19}. electrons flow per second.

<h3>What causes a current in a wire?</h3>
  • Electric current in a wire, where electrons serve as the charge carriers, is a measurement of the amount of charge that moves through any point of the wire in a given amount of time.
  • A free electron is drawn to a proton to become neutral if an electron is added to the wire.
  • Lack of electrons can result from pushing electrons out of their orbits.
  • Electric current is the name given to the constantly moving electrons in wire.

The current is the quantity of charge Q flowing through a certain point of the wire in a time interval of \Delta t.

I = \frac{Q}{\Delta t}.

by using this relationship

I=1.80 A, we can find the charge passing any point in the wire in 1 second:

Electric Charge, Q = 1.80 C.

To find how many electrons corresponds to this charge, we should divide this value by the charge of a single electron

charge of the electron = 1.6 x 10^{-19} C.

No. of Electrons = Q/q = \frac{1.80}{1.6* 10^{-19}}= 1.12 x 10^{19}.

To learn more about Electric current   refer,

brainly.com/question/9467901

#SPJ4

4 0
1 year ago
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