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lana [24]
2 years ago
14

A 150-loop coil of cross sectional area 6.3 cm2 lies in the plane of the page. An external magnetic field of 0.060 T is directed

out of the plane of the page. The external field decreases to 0.030 T in 15 milliseconds.
(a) What is the magnitude of the change in the external magnetic flux enclosed by the coil
(B) What is the magnitude of the average voltage induced in the coil as the external flux is changing?
(C) If the coil has a resistance of 4.0 ohms, what is the magnitude of the average current in the coil?
Physics
1 answer:
algol [13]2 years ago
3 0

Answer:

Explanation:

area of loop = 6.3 x 10⁻⁴ m²

Total flux ( initial )

nB A

=  150 x .06 x  6.3 x 10⁻⁴

Final magnetic flux

150 x .03 x  6.3 x 10⁻⁴

change in flux

150 x .06 x  6.3 x 10⁻⁴ - 150 x .03 x  6.3 x 10⁻⁴

= 28.35 x 10⁻⁴ Weber .

b ) voltage induced = rate of change of flux

= change in flux / time

= 28.35 x 10⁻⁴  / 15 x 10⁻³

= 1.89 x 10⁻¹ V

c )

current induced = voltage induced / resistance

= 1.89 x 10⁻¹  / 4

= .4725 x 10⁻¹ A

= 47.25  m A .

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A 90 kg astronaut Travis is stranded in space at a point 12 m from his spaceship. In order to get back to his ship, Travis throw
insens350 [35]

Answer:

Explanation:

This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:

[m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a which, in words, is

The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:

[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a

Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).

0 = 90.0v - 2.0 and

2.0 = 90.0v so

v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.

12 = .022t and

t = 550 seconds, which is the same thing as 9.2 minutes

7 0
3 years ago
5/137 Under the action of its stern and starboard bow thrusters, the cruise ship has the velocity vB = 1 m/s of its mass center
quester [9]

The image is missing, so i have attached it;

Answer:

A) V_rel = [-(2.711)i - (0.2588)j] m/s

B) a_rel = (0.8637i + 0.0642j) m/s²

Explanation:

We are given;

the cruise ship velocity; V_b = 1 m/s

Angular velocity; ω = 1 deg/s = 1° × π/180 rad = 0.01745 rad/s

Angular acceleration;α = -0.5 deg/s² = 0.5 x π/180 rad = -0.008727 rad/s²

Now, let's write Velocity (V_a) at A in terms of the velocity at B(V_b) with r_ba being the position vector from B to A and relative velocity (V_rel)

Thus,

V_a = V_b + (ω•r_ba) + V_rel

Now, V_a = 0. Thus;

0 = V_b + (ω•r_ba) + V_rel

V_rel = -V_b - (ω•r_ba)

From the image and plugging in relevant values, we have;

V_rel = -1[(cos15)i + (sin15)j] - (0.01745k * -100j)

V_rel = - (cos15)i - (sin15)j - 1.745i

Note that; k x j = - i

V_rel = [-(2.711)i - (0.2588)j] m/s

B) Let's write the acceleration at A with respect to B in terms of a_b.

Thus,

a_a = a_b + (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_a and a_b = 0.

Thus;

0 = (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_rel = - (α*r_ba) - (ω(ω•r_ba)) - (2ω*v_rel)

Plugging in the relevant values with their respective position vectors, we have;

a_rel = - (-0.008727k * -100j) - (0.01745k(0.01745k * -100j)) - (2*0.01745k * [-(2.711)i - (0.2588)j])

a_rel = 0.8727i - (0.01745² x 100)j + 0.0946j - 0.009i

Note that; k x j = - i and k x i = j

Thus,simplifying further ;

a_rel = 0.8637i + 0.0642j m/s²

4 0
3 years ago
A student wants to design an experiment to study the transformation of mechanical energy. Which object can be used to investigat
Vsevolod [243]
I guess it’s the 3rd one
7 0
2 years ago
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There is a plate with moment of inertia Ip = 0.0711 kgm2 , rotating around its center of mass with angular speed of 4.53 rad/s.
Anna007 [38]

Answer:

1) their common angular speed = 3.038 rad/s

2) kinetic energy loss = 0.24J

Explanation:

1) This is a case of conservation of angular momentum.

The initial angular momentum must be equal to the final angular momentum

Initial angular momentum = I x w

Where I = moment of inertia, and

w = angular momentum.

Initial angular momentum = 0.0711 x 4.53 = 0.322 kg-m2-rad/s

After addition of ring, moment of inertia becomes,

0.0711 + 0.0353 = 0.106 kg-m2

Therefore, final angular momentum will be

0.106 x Wf

Where Wf = final common angular velocity

Equating the two angular momentum we have

0.322 = 0.106Wf

Wf = 0.322/0.106 = 3.038 rad/s

2) KE = 1/2 x I x w^2

Initial KE = 1/2 x 0.0711 x 4.53^2

= 0.729 J

Final KE = 1/2 x 0.106 x 3.038^2

= 0.489 J

Loss in KE = 0.729 - 0.489 = 0.24 J

5 0
3 years ago
Create a list of 5 potential jobs that students of genealogy can obtain.
Kipish [7]

Answer:

Private Investigator.

Investigative Genetic Genealogist.

Historic Preservationist.

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Heir Searcher.

Explanation:

4 0
2 years ago
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