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Nutka1998 [239]
4 years ago
8

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 11.8 cm^2 is rotated during the time interval 4.9

0×10^-2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.60×10-5 T.
a) What is the total magnitude of the magnetic flux through the coil before it is rotated?


b) What is the magnitude of the total magnetic flux through the coil after it is rotated?


c) What is the magnitude of the average emf induced in the coil?
Physics
1 answer:
Zanzabum4 years ago
6 0

Answer:

Explanation:

Flux through the coil = nBA , n is no of turns , B is magnetic flux and A a is area of the coli

= 200 x 5.6 x 10⁻⁵ x 11.8 x 10⁻⁴

=  13216 x 10⁻⁹ weber .

b ) When the coil becomes parallel to magnetic field  , flux through it will become zero.

c ) e m f induced = change in flux / time

= 13216 x 10⁻⁹ / 4.9 x 10⁻²

= 2697.14 x 10⁻⁷ V

= 269.7 x10⁻⁶

269.7 μV.

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NASA launches a satellite into orbit at a height above the surface of the Earth equal to the Earth's mean radius. The mass of th
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Answer:

a) 3h 59’  b) 5,590 m/s  c) 1,910 N

Explanation:

The only force acting on the satellite (neglecting the influence of any other body) is just the attractive force from Earth, which is given by the Universal Law of Gravitation:

Fg = G* ms*me / (res² (1)

where:  G = 6.67* 10⁻¹¹ N.m² / kg² ms = 780 kg, me= 5.97* 10²⁴ kg, and  

res = 2* 6.38* 10⁶ m (distance between the center of the Earth and the satellite).

Fg = 1,910 N (answer c)

At the same time, this force, is the centripetal force that keeps the satellite in orbit, and that can be written as follows:

Fg = ms * v² / res (2)

By definition of velocity, we can say that the constant speed at which the satellite orbits, can be expressed as the quotient between the distance travelled around the Earth once, and the time needed to do that, which is called the period of the orbit:

v = 2*π*res / T (3)

We can solve for v first, taking the right sides of (1) and (2), as follows:

G* ms*me / (res)²= ms * v² / res

v = √(G*me/r) = 5,590 m/s (answer b)

Once obtained the value of v, we can replace in (3), and solve for T:

T = 2* π*res / v = 3 h 59 min (answer a)

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3 years ago
An advertisement uses a movie star holding a bottle of perfume. This ad is an example of which of the following?
icang [17]

The Answer would be (A) an appeal of association

7 0
3 years ago
S the work done on Amanda's car while speeding up (i) greater than, (ii) less than, or (iii) the same as the work done on Bertha
ankoles [38]

Answer:

Explanation:

Mass of amanda and bertha car = Ma = Mb

Initial velocity of amanda, ua = 10 m/s

Final velocity of amanda, va = 20 m/s

Initial velocity of bertha, ub = 20 m/s

Final velocity of bertha, vb = 30 m/s

Workdone = change in Energy = 1/2 mv^2 - 1/2 mu^2

Ea = Ma × 1/2 × (20^2 - 10^2)

= Ma × 150

= 150 Ma J

Workdone = change in Energy = 1/2 mv^2 - 1/2 mu^2

Eb = Mb × 1/2 × (30^2 - 20^2)

= Mb × 250

= 250 Mb J

Option ii. Less than. Thus this is because the workdone by amanda car (150Ma) is less than the workdone in berthas car (250Mb).

B.

Impulse, p = force × time

p = Mass × change in velocity

pa = Ma × va - Ma × ua

= Ma × (20 - 10)

= 10 × Ma

= 10 Ma

p = Mass × change in velocity

pa = Mb × vb - Mb × ub

= Mb × (30 - 20)

= 10 × Mb

= 10 Mb

Option iii. The same as.

The impulse as seen above is the same in amanda car (10Ma) as the same with bertha (10Mb)

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Temperature differences cause pressure differences which ultimately causes
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Answer:

...

Explanation:

...

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4 years ago
If the height is 10m, what is the potential energy? Asuming that the mass is 10
Ratling [72]

\huge{\textit{\sf{{\color{m}{Answer}}{\color{b}{࿐}}}}}

we know,

\boxed{potential \: energy = mgh}

So,

\longmapsto10 \times 9.8 \times 10

\longmapsto980 \: joules

6 0
3 years ago
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