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Ne4ueva [31]
3 years ago
11

A horizontail rod (oriented in the east -west direction) is moved northward at a constant velocity through a magnetic field that

points straight down. Make a statement concerning the potential induced across the rod.
A) The east end of the rod is at higher potential than the west end.
B) The bottom surface of the rod is at higher potential than the top surface.
C) The top surface of the rod is at higher potential than the bottom surface.
D) The west end of the rod is at higher potential than the east end.
E) The potential is uniform.
Physics
1 answer:
Ede4ka [16]3 years ago
6 0

Answer:

a

Explanation:

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A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f
Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

T=\dfrac{2\pi}{\omega}

{\omega}=\dfrac{2\pi }{0.2}\ rad/s

{\omega}=31.41\ rad/s

We know that

{\omega}^2=m\ K

K=Spring constant

K=\dfrac{\omega^2}{m}

K=\dfrac{31.41^2}{0.12}\ N/m

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0
4 years ago
An 4-kg ball experiences a force and accelerates at a rate of 1.5 m/s.
ANTONII [103]

Answer:

6.0 N

Explanation:

The strength of a force is expressed as the magnitude of the force in Newton.

The formula to apply here is :

Force= mass * acceleration

F=ma

Mass, m = 4 kg

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Force= 4 *1.5 = 6.0 N

4 0
3 years ago
A guitarist finds that the pitch of one of her strings is slightly flat—the frequency is a bit too low. Should she increase or d
Yuri [45]

Answer:

The guitarist should increase the tension of the string.

Explanation:

The frequency of a vibrating string is determined by fn=(n/(2L))√T/μ. So if the tension in the string increased, the rate at which it vibrates (the frequency) will also increase.

Therefore it is advisable that she increase the tension of the string.

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</span>I is the current passing (I = 4 amperes)
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Substituting in the mentioned equation, we find that:
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