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ExtremeBDS [4]
3 years ago
13

Lisa made the electromagnet shown. A nail with wire coiled around it has its head labeled S to the right and its point labeled N

to the left. The end of the wire leading to the S is attached to the positive terminal of a battery. The end of the wire leading to the N is attached to the negative terminal of the battery. What can Lisa do to increase the strength of the electromagnet? She can use a nail with weaker magnetic properties. She can change the direction of the nail. She can increase the number of wire loops. She can reduce the current in the wire.
Physics
2 answers:
Assoli18 [71]3 years ago
7 0

Answer:

She should definitely increase the number of coils around the nail.

Bogdan [553]3 years ago
5 0

Answer:

C. She can increase the number of wire loops.

Explanation:

The more wire loops the more energy.

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Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus
r-ruslan [8.4K]

Answer:

nπ + π/2 for any integer n

Explanation:

Since destructive interference occurs every odd multiple of half wavelength, that is π/2, 3π/2, 5π/2 where the interference is half wavelength and in general, (n + 1/2)π where n is an integer. So, nπ + π/2 for any integer n

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An object is held at an unknown height above Earth’s surface, where the acceleration due to gravity of the object is considered
Aleks04 [339]

Answer:

Options A and B.

Explanation:

Gravitational acceleration, initial height, intial speed and time are required to determine final speed. The option D is incorrect, since speed varies in time. Option C is dimentionally wrong.

The correct strategy is calculating the initial height from option B. Later, substituting time in equation A to derive an expression of the final velocity in terms of position. Hence, the required equations are options A and B.

7 0
3 years ago
[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the
ExtremeBDS [4]

Answer:

t = 5.89 s

Explanation:

To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.

Let us assume that the radius of the pulley (r_p) = 0.4 m

Let the radius of the sphere (r) = 0.5 m

w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s

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mass (m) = 3 kg each

\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw

t = \frac{4r^2mw}{Tr_P}

Substituting values:

t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s

7 0
3 years ago
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Answer:

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As the air resistance be negligible, is time taken to reach the ground is also 2 sec.

So, total time taken be the ball to reach at original point = 2 + 2 = 4 s

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