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svetlana [45]
2 years ago
7

A circular loop of wire is rotated at constant angular speed about an axis whose direction can be varied. In a region where a un

iform magnetic field points straight down, what must be the orientation of the loop's axis of rotation if the induced emf is to be zero?
Physics
1 answer:
katovenus [111]2 years ago
5 0

Answer:

here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field

Explanation:

As we know by Faraday's law of electromagnetic induction

Rate of change in magnetic flux will induce EMF in the coil

so here we will have

EMF = \frac{d\phi}{dt}

here we know that

\phi = NB.A

now if the magnetic flux will change with time then it will induce EMF in the coil

EMF = N\frac{d}{dt}(B.A)

so here induced EMF will be zero in the coil if the flux linked with the coil will remain constant

so here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field

In such a way when coil will rotate then the flux linked with the coil will remains constant and there will be no induced EMF in it

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A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga
andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

We know that heat addition at constant volume process ( rigid vessel ) given as

Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

Cp-Cv= R

Cp=7/2 R

Now by putting the values

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

Process is adiabatic

\Delta S_{surr}=0

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

\Delta S_{univ}= 20.79 +0

\Delta S_{univ}= 20.79

3 0
2 years ago
Which formula is used to find fluctuation of the shape of body
Sladkaya [172]

Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

5 0
2 years ago
Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by th
mote1985 [20]

Answer:

U = 1794.005 × 10⁶ J

Explanation:

Data provided;

Capacitance of the original capacitor, C = 1.27 F

Potential difference applied to the original capacitor, V = 59.9 kV

= 59.9 × 10³ V

Now,

The Potential energy (U) for the capacitor is calculated as:

Potential energy of the original capacitor, U = \frac{\textup{1}}{\textup{2}}  × C × V²

on substituting the respective values, we get

U = \frac{\textup{1}}{\textup{2}}  × 1.27 × ( 59.9 × 10³ )²

or

U = 1794.005 × 10⁶ J

7 0
2 years ago
What is the net force required to give an automobile with a mass of 1,600 kg an acceleration of 4.5 m/s2
Paladinen [302]

Answer:

Fnet=7200 N

Explanation:

Fnet=mass x acceleration

mass= 1600kg acceleration=4.5m/s^2

8 0
2 years ago
Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their
Vera_Pavlovna [14]

Answer:

A = 2.36m/s

B = 3.71m/s²

C = 29.61m/s2

Explanation:

First, we convert the diameter of the ride from ft to m

10ft = 3m

Speed of the rider is the

v = circumference of the circle divided by time of rotation

v = [2π(D/2)]/T

v = [2π(3/2)]/4

v = 3π/4

v = 2.36m/s

Radial acceleration can also be found as a = v²/r

Where v = speed of the rider

r = radius of the ride

a = 2.36²/1.5

a = 3.71m/s²

If the time of revolution is halved, then radial acceleration is

A = 4π²R/T²

A = (4 * π² * 3)/2²

A = 118.44/4

A = 29.61m/s²

7 0
2 years ago
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