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riadik2000 [5.3K]
2 years ago
14

A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18

m/s in the x-direction, 24 m/s in the y-direction, and 72 m/s in the z-direction. The segment has length 0.50 m and is parallel to the z-axis as it moves.Part1) Find the motional emf induced between the ends of the segment.
Part 2) What would the motional emf be if the wire segment was parallel to the y-axis?
Physics
1 answer:
Annette [7]2 years ago
6 0

Answer:

Part a)

Induced EMF when length vector is along Z direction is 0.72 V

Part b)

Induced EMF when length vector is along Y direction is ZERO

Explanation:

As we know that the motional EMF induced in the wire is given as

E = (v\times B). L

1)

As we know that

v = 18\hat i + 24\hat j + 72\hat k

B = 0.080 \hat j

L = 0.50 \hat k

now we have

\vec v \times \vec B = 1.44\hat k - 5.76 \hat i

so we have

E = 1.44 (0.50) = 0.72 V

2)

If the length vector is along Y direction then we have

L = 0.50 \hat j

so again we have

\vec v \times \vec B = 1.44\hat k - 5.76 \hat i

so we have

EMF = 0

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A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration
wel

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

V=X/T

First stage:

T1=5s

v_{f}  =v_{o} - at

But, v_{f}  =0   (decelerates to rest)

then: a =v_{o} /t=0.3/5=0.06m/s^{2}

on the other hand:

x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m

X1=75cm

Second stage:

T2=5s

x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

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2 years ago
A typical human contains 5.00 l of blood, and it takes 1.00 min for all of it to pass through the heart when the person is resti
Mrrafil [7]
<span>(a) 0.0676 l (b) 67.6 cc So we've been told that 5.00 L of blood flows through the heart every minute and that the heart beats 74.0 times per minute. So that means that for every beat of the heart, 5.00 L / 74.0 = 0.067567568 L of blood flows through the heart. Rounding to 3 significant figures gives 0.0676 l. Converting from liters to cubic centimeters simply require a multiplication by 1000, so we have 67.6 cc of blood pumped per beat.</span>
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2 years ago
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Whale sharks swim forward while ascending or descending. They swim along a straight-line path at a shallow angle as they move fr
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You did not include the quesetion, but I can help you to understand the problem and how to find the relevant information.

1) The angle of 13° with which the shark ascends meets this:

Vertical ascending velocity = 0.85m/s * sin(13°)

Horizontal velocity = 0.85m/s * cos(13°)

2) The length swan by the shark ascending meets this

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Horizontal length, y:

\frac{y}{50} = \frac{0.85sin(13)}{0.85cos(13)}

From that y = 50 * tan(13°)

=> y = 11.54 m.

3) Conclusions:

1) The shark run 50 m vertically upward and 11.54 m horizontally.

2) The length of the path run by the shark may be calculated using Pythagoras' theorem:

hypotenuse^2 = (50m)^2 + (11.54m)^2 = 2633.25m^2

hypotenuse = 51.35m

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2 years ago
Jill puts her face in front of a convex mirror, 18 cm from the focal point of the mirror. If the focal point is located 12 cm fr
Bumek [7]
A. 4 cm behind the mirror
<span>  For any mirror, </span><span><span>so</span><span>si</span>=<span>f^2</span></span>. Therefore, by plugging in the values, you get <span>18<span>si</span>=144. 144/18 = 8, so </span><span><span>si</span>=8</span>
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ivolga24 [154]
1) Focal length

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\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i} (1)
where 
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d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

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from which we find
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3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
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