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2 years ago

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In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil 1, which has

228 loops, is 3.13 V. The emf induced in coil 2 is 4.16 V. How many loops does coil 2 have

1 answer:

Nat2105 [25]2 years ago

5 0

Answer:

303

Explanation:

We are given that

Emf in coil 1, $E_1=3.13 V$

Emf induced in coil 2, $E_2=4.16 V$

Number of loops in coil 1, $N_1=228$

We have to find the number of loops in coil 2.

Rat of change of magnetic flux in a single loop is same.

Let $\phi_1$ and $\phi_2$ be the magnetic flux in coil 1 and coil 2.

$\frac{d\phi_1}{dt}=\frac{d\phi_2}{dt}$

$\frac{V_2}{V_1}=\frac{N_2}{N_1}$

Using the formula

$\frac{E_2}{E_1}=\frac{N_2}{N_1}$

$\frac{4.16}{3.13}=\frac{N_2}{228}$

$N_2=\frac{4.16}{3.13}\times 228$

$N_2=303$

Hence, the number of loops in coil 2=303

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Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the

IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency $f= 6.37\times10^{14}\ Hz$

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{c}{f}$

where, c = speed of light

f = frequency

Put the value into the formula

$\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}$

$\lambda=471\ nm$

We need to calculate the distance between the two slits

$m\times \lambda=d\sin\theta$

$d =\dfrac{m\times\lambda}{\sin\theta}$

Where, m = number of fringe

d = distance between the two slits

Here, $\sin\theta =\dfrac{y}{l}$

Put the value into the formula

$d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}$

$d=40.11\times10^{-6}\ m$

$d = 40.11\ \mu m$

Hence, The distance between the two slits is 40.11 μm.

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3 years ago

What is the rock cycle and how does it change the lithosphere?

Ira Lisetskai [31]

The rock cycle is a basic concept in geology that describes the time-consuming transitions through geologic time among the three main rock types: sedimentary, metamorphic, and igneous. As the adjacent diagram illustrates, each of the types of rocks is altered or destroyed when it is forced out of its equilibrium conditions. An igneous rock such as basalt may break down and dissolve when exposed to the atmosphere, or melt as it is subducted under a continent. Due to the driving forces of the rock cycle, plate tectonics and the water cycle, rocks do not remain in equilibrium and are forced to change as they encounter new environments. The rock cycle is an illustration that explains how the three rock types are related to each other, and how processes change from one type to another over time. This cyclical aspect makes rock change a geologic cycle and, on planets containing life, a biogeochemical cycle.

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sources: wikapedia, Harmonybaddie on brainly

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3 years ago

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert

cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, $k=4.2\times 10^5\ N/m$

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

$mgh=\dfrac{1}{2}kx^2$

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So, the compression in the spring is 5.88 meters.

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3 years ago

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Ede4ka [16]

Answer:

B

Explanation:

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3 years ago

Read 2 more answers

Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp

tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s 2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

½ *m*v² + ½* I* ω² = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

Solid Sphere I = 2/5* m *r²

Replacing in (1):

½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(10/7*9.8 m/s2*14.7 m) = 14. 4 m/s

Spherical shell I=2/3*m*r²

Replacing in (1):

½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(6/5*9.8 m/s2*14.7 m) = 13.2 m/s

Hoop I= m*r²

Replacing in (1):

½ * v² + ½ (m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(9.8 m/s2*14.7 m) = 12.0 m/s

Cylinder I = 1/2 * m* r²

Replacing in (1):

½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

Replacing by the value given for h, and solving for v:

v = 2*√(1/3*9.8 m/s2*14.7 m) = 13.9 m/s

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3 years ago