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kobusy [5.1K]
2 years ago
15

In a certain time period a coil of wire is rotated from one orientation to another with respect to a uniform 0.38-T magnetic fie

ld. The emf induced in the coil is 5.8 V. Other things being equal, what would the induced emf be if the field had a magnitude of 0.55 T?
Physics
1 answer:
Evgesh-ka [11]2 years ago
4 0

Answer:

8.4 V

Explanation:

induced emf, e1 = 5.8 V

Magnetic field, B1 = 0.38 T

magnetic field, B2 = 0.55 T

induced emf, e2 = ?

As we know that the induced emf is directly proportional to the magnetic field strength.

When the other parameters remains constant then

\frac{e_{1}}{e_{2}}=\frac{B_{1}}{B_{2}}

\frac{5.8}{e_{2}}=\frac{0.38}{0.55}

e2 = 8.4 V

Thus, the induced emf is 8.4 V.

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Why is the term minority group no longer effective
Cloud [144]
The term minority group is no longer effective because these groups now make up significant percentages of the total population
8 0
3 years ago
A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, ca
viva [34]

Answer:

a=0.2*10^{-5}g

Explanation:

From the question we are told that:

Mass M=180=>0.18kg

Charge Q=18mC=18*10^-^3C

Velocity v=2.2m/s

Length of Wire L=8.6cm=>0.086

Current I=30A

Generally the equation for Magnetic Field of Wire B is mathematically given by

 B=\frac{\mu_0*I}{2\pi*l}

 B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}

 B=6.978*10^{-5}T

Generally the equation for Force on the plane F is mathematically given by

 F=qvB

Therefore

 ma=qvB

 a=\frac{qvB}{m}

 a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}

 a=2.37*10^{-5}

Therefore in Terms of g's

 a=\frac{2.37*10^{-5}}{9.8}

 a=0.2*10^{-5}g

8 0
2 years ago
If the acceleration of a motorboat is 4.0 m/s2, and the motorboat starts from rest, what is its velocity after 6.0 s?
umka2103 [35]

Answer:

The velocity of the motorboat after 6s is 24 m/s.

Explanation:

Given;

acceleration of the motorboat, a = 4.0 m/s²

initial velocity of the motorboat, u = 0

time of motion of the motorboat = 6s

Apply the following kinematic equation to determine the velocity of the motorboat after 6 ;

v = u + at

v = 0 + (4 x 6)

v = 24 m/s

Therefore, the velocity of the motorboat after 6s is 24 m/s.

6 0
3 years ago
Example: A wooden crate with mass 100kg is at rest on a stone floor. You know that the coefficients of kinetic and static fricti
alexgriva [62]

Answer

Any force greater 490N

Explanation

The force required just to make an object slide over a rough horizontal surface is any force greater that the static friction which given by;

F=\mu_s mg.............(1)

Given;

\mu_s=0.5\\m=100kg\\g=9.8m/s^2

Hence;

F = 0.5 x 100 x 9.8

F = 490N.

We will only need the coefficient of kinetic friction if we were asked to find the force required to keep the object moving uniformly. Usually, the force needed to keep an object moving uniformly over a rough surface is lesser that which is needed to start its motion.

In this problem, we were only asked to find the minimum force required to make the object move which we have done.

7 0
3 years ago
Calculate the pressure exerted on the ground by a boy of a mass 60 kg if he stands on one foot.the area of the sole of his shoe
ddd [48]

Answer:

40 Kpa

Explanation:

150 cm2 = 0.015 m2

p \:  =  \frac{mg}{ a}  = 40000

8 0
2 years ago
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