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3 years ago

Find the magnitude of the emf induced in the rod when it is moving toward the right with a speed 7.50 m/s.

Physics

1 answer:

Flauer [41]3 years ago

6 0

Answer:

3 volts

Explanation:

It is given that,

Magnetic field, B = 0.8 T

Length of a conducting rod, l = 50 cm = 0.5 m

Velocity of the conducting rod, v = 7.5 m/s

We need to find the magnitude of the emf induced in the rod when it is moving toward the right. When a rod is moved in a magnetic field, an emf is induced in it and it is given by :

$\epsilon=Blv$

Putting all the values,

$\epsilon=0.8\times 0.5\times 7.5\\\\\epsilon=3\ V$

So, the magnitude of the emf induced in the rod is 3 volts.

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A 2 kg car accelerated from 10 m's to 20 m/s using a force of 3000N. How quickly did it

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Answer:

the car accelerates in

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Explanation:

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3 years ago

A baseball is thrown at an angle of 40.0° above

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Explanation:

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3 years ago

The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

$N_A+N_B = 45*9.8$

$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :

Then:

$\sum M _B = 0$

$P*h = mg*0.24$

$h =\frac{45*9.8*0.24}{132.3}$

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

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