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UkoKoshka [18]
2 years ago
6

An 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its

plane is parallel to the field direction in 0.20 s. What is the average induced emf in the loop?
Physics
1 answer:
nikdorinn [45]2 years ago
4 0

Answer:

The average induced emf in the loop is 0.20 V

Explanation:

Given:

Radius of loop r = \frac{d}{2} = 9.25 \times 10^{-2} m

Magnetic field B = 1.5 T

Change in time \Delta t = 0.20 sec

According to the faraday's law,

Induced emf is given by

    \epsilon = -\frac{\Delta \phi}{\Delta t}

Where \phi = magnetic flux

  \phi = BA\cos0                                ( here \theta = 0 )      

Where A = \pi r^{2}

We neglect minus sign because it's shows lenz law

   \epsilon  = \frac{B \pi r^{2}  }{\Delta t}

   \epsilon = \frac{1.5 \times 3.14 \times (9.25 \times 10^{-2} )^{2} }{0.20}

   \epsilon = 0.20 V

Therefore, the average induced emf in the loop is 0.20 V

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Determine the sign (+ or −) of the torque about the elbow caused by the biceps, τbiceps, the sign of the weight of the forearm,
Alex Ar [27]
Ans: 
1.  τbiceps = +(Positive)
2.  τforearm = -(Negative)
3.  τball = -(Negative)

Explanation:

The figure is attached down below.

1. T<span>orque about the elbow caused by the biceps, τbiceps:
Since Torque = r x F (where r and F are the vectors)
</span>Where r is the vector from elbow to the biceps.
<span>
We can see in the figure that F(biceps) is in upward direction, and by applying the right hand rule from r to F, we get the counterclockwise direction. The torque in counterclockwise direction is positive(+). Therefore, the sign would be +.

2. </span>Torque about the the weight of the forearm, τforearm:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the forearm.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(forearm) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

3. Torque about the the weight of the ball, τball:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the ball.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(ball) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

8 0
2 years ago
You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle
charle [14.2K]

Answer:

a=5.54m/s^{2}

Explanation:

The net force, F_{net} of the box is expressed as a product of acceleration and mass hence

F_{net}=ma where m is mass and a is acceleration

Making a the subject, a= \frac {F_{net}}{m}

From the attached sketch,  

∑ F_{net}=Fcos\theta-F_{f} where F_{f} is frictional force and \theta is horizontal angle

Substituting ∑ F_{net} as F_{net} in the equation where we made a the subject

a= \frac {Fcos\theta-F_{f}}{m}

Since we’re given the value of F as 240N, F_{f} as 41.5N, \theta as 30^{o} and mass m as 30kg

a= \frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}

6 0
2 years ago
A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h
Brilliant_brown [7]

Answer:

Explanation:

Impulse = change in momentum

mv - mu , v and u are final and initial velocity during impact at surface

For downward motion of baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes final velocity after impact

change in momentum

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

= .7056 N.s.

Impulse by floor in upward direction

= .7056 N.s

6 0
2 years ago
Suzette had prepared the graph below to add to her lab
Charra [1.4K]

Answer:

A title

Explanation:

Because this is middle school.

4 0
2 years ago
Read 2 more answers
Two cars start 200 m apart and drive toward each other at a steady 10 m/s. On the front of one of them, an energetic grasshopper
vladimir1956 [14]

Answer:

Total distance does the grasshopper travel before the cars hit is 150 m

Explanation:

Each car moves x=100 m before they collide. Both the cars moving in constant velocity. time taken t by each car is

t=\frac{x}{v}

where x  is the distance traveled with velocity v

t=\frac{100}{10}\\t=10 sec

The insect is moving through this time period with a constant velocity of 15 m/s

The distance traveled by grasshopper  is

distance=V_{gh} \times t\\distance=15 \times 10\\distance=150 m

7 0
2 years ago
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