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Tomtit [17]
4 years ago
5

A 0.45-m metal rod moves 0.11 m in a direction that is perpendicular to a 0.80-T magnetic field in an elapsed time of 0.036 s. A

ssuming that the acceleration of the rod is zero m/s2, determine the emf that exists between the ends of the rod.
Physics
2 answers:
madam [21]4 years ago
6 0

Answer:

(B) This cannot be determined without knowing the orientation of the rod relative to the magnetic field.

Because, Since acceleration of the rod is Zero. So, net force acting on the rod will be zero. But, its not specified that whether the rod is moving along varying magnetic field direction or a constant magnetic field direction.

Gwar [14]4 years ago
3 0

Answer:

1.1 V

Explanation:

L = 0.45 m

d = 0.11 m

B = 0.80 T

t = 0.036 s

Let e be the emf.

e = B v L

e = 0.80 x 0.11 x 0.45 / 0.036 = 1.1 V

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Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

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You are loading a refrigerator weighing 2267 N onto a truck, using a wheeled cart. The refrigerator is raised 1.09 m to the truc
meriva

Answer:

a).

Wmin= 2471.03 J

W=1603.01 J

Explanation:

Weight, w= 2267 N

w= m*g\\W=m*g*h \\W=w*h

Minimum work 'h' is the distance the refrigerator is raised h=1.09m

W_{min}=2267 N* 1.09m\\W_{min}=2471.03 J

The motion is no frictional force so, the magnitude of the force with a angle of 45.0° is find using:

W=m*g*h'\\h'= sin(45)\\W=2267N*sin(45)\\W=1603.01 J

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A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th
solniwko [45]

Answer:

a = R\alpha\sqrt{1 + \alpha^2t^4}

Explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

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2) Centripetal acceleration

a_c = \omega^2 R

here we know that

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now we know that net linear acceleration is given as

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so we have

a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}

a = R\alpha\sqrt{1 + \alpha^2t^4}

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3 years ago
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