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A coil of wire with 50 turns lies in the plane of the page and has an initial area of 0.180 m2. The coil is now stretched to hav

butalik [34]

Answer:

Explanation:

Given that,

Number of turns of coil

N = 50 turns

Initial area of plane

A1 = 0.18 m²

The coil it stretch to a no area in time t = 0.1s

No area implies that the final area is 0, A2 = 0 m²

Constant magnetic field strength

B = 1.51 T

EMF?

EMF is given as

Using far away Lenz law

ε = —N• dΦ/dt

Where Φ = BA

Then,

ε = —N• d(BA)/dt

Since B is constant,

ε = —N•B dA/dt

ε = —N•B (∆A/∆t)

ε = —N•B(A2—A1)/(t2-t1)

ε = —50 × 1.51 (0—0.18)/(0.1—0)

ε =—75.5 × —0.18 / 0.1

ε = 135.9 V

The induced EMF is 135.9V

Fleming’s left hand rule stated that if the index finger points toward magnetic flux, the thumb towards the motion of the conductor, then the middle finger points towards the induced emf.

Since the area lines in the plane, then the induced emf will be out of the page

5 0

4 years ago

Refraction occurs when a wave enters a new medium at an angle because

devlian [24]

Answer

The refracted wave must obey Snell's equation

Ni sin θi = Nr sin θr

If Nr differs from Ni then sin θr will differ from sin θi

If the wave originates from point A and ends at point B then Snell's Law shows that the time for light to get from point A to point B is a minimum.

4 0

3 years ago

A beam of light passes from air into a cube of zirconium if the angle of incidence is 40”what is the angle of refraction ?

Colt1911 [192]

Answer:

50 degrees

Explanation:

Usually in physics in solving such questions refractive index of air is taken as 1.00 (n=1.00) by assuming that air is a vacuum. Hence, the light ray passes from water to air at an angle of incident of 40 degrees, it means that it has angle of 50 degrees to the normal line.

5 0

4 years ago

In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p

Bond [772]

Answer:

$B_0 = 1.69 \times 10^{-4}\ T$

Explanation:

given,

total deflection = 4.12 cm

Electric field = 1.1 ×10³ V/m

plate length = 6 cm

distance between them = 12 cm

using formula

$v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}$

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

L = 0.12 m

$v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}$

v_0 = 6496355.63 m/s

$v_0 = \dfrac{E}{B_0}$

$B_0 = \dfrac{E}{v_0}$

$B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}$

$B_0 = 1.69 \times 10^{-4}\ T$

5 0

3 years ago

g A uniform ladder whose length is 4.4 m and whose weight is 330 N leans against a frictionless vertical wall. The coefficient o

Damm [24]

Answer:

Distance = 4.4 [m]

Explanation:

This problem can be easily solved using a static analysis of forces acting on the ladder, taking into account the respective distances. For easy understanding, a free body diagram should be made.

We perform a sum of force on the X-axis equal to zero, to find that the force exerted by the wall is equal to the friction force on the floor.

Then we perform a summation of forces on the Y axis, to determine that the normal force exerted by the floor is equal to the weight of the ladder.

We know that the friction force is equal to the product of normal force by the coefficient of friction.

In this way, by relating the friction force to the equations deduced above we can find the force exerted by the wall.

Then we make a summation of moments around the base point of the ladder, the equation realized can be seen in the attached image.

In the last analysis we can find the relationship between the horizontal and vertical distance of the ladder, with respect to the wall and the floor.

Then with the complementary analysis of the Pythagorean theorem we can find another additional equation.

The result of the greater distance is 4.4 [m]

3 0

3 years ago

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