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xenn [34]
2 years ago
15

A technician is working on an MRI machine. To test it, the technician turns on the MRI machine that produces a strong magnetic f

ield, but then he accidentally drops a screwdriver in this magnetic field. As the screwdriver falls, an emf is induced across the length of the metal shaft, L, of the screwdriver. At one instant, the screwdriver falls with a speed of 2.1 m/s and is oriented as shown in figure so that the induced emf is maximized. The magnitude of the magnetic field at this instant is 2.5 T, and the length of the metal shaft, L, is 10 cm. What is the induced emf across the length of the metal shank at this instant?
Physics
1 answer:
AnnZ [28]2 years ago
4 0

Answer:

0.525 V

Explanation:

Noting the given data in the question:

we have,

The speed of the fall of the screw driver, v = 2.1 m/s

The magnitude of the magnetic field, B = 2.5 T

The length of the metal shaft of the screw driver, L = 10 cm = 0.1 m

Now, the induce emf (E) is given as:

E = B × L × v

on substituting the values in the above equation, we get

E = 2.5 × 0.1 × 2.1

or

E = 0.525 V

Hence, the emf induced across the length of the metal shank will be 0.525 V

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. A girl runs and jumps horizontally off a platform 10m above a pool with a speed of 4.0m/s. As soon as she leaves the platform,
faust18 [17]

Answer:

2.39 revolutions

Explanation:

As she jumps off the platform horizontally at a speed of 10m/s, the gravity is the only thing that affects her motion vertically. Let g = 10m/s2, the time it takes for her to fall 10m to water is

h = gt^2/2

10 = 10t^2/2

t^2 = 2

t = \sqrt{2} = 1.414 s

Knowing the time it takes to fall to the pool, we calculate the angular distance that she would make at a constant acceleration of 15 rad/s2:

\theta = \alpha t^2/2

\theta = 15 * 2/2 = 15 rad

As each revolution is 2π, the total number of revolution that she could make is: 15 / 2π = 2.39 rev

3 0
2 years ago
Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele
Julli [10]

Answer:

0.018 J

Explanation:

The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

W = q \Delta V

where

q=3.0 \mu C = 3.0 \cdot 10^{-6} C is the magnitude of the charge

\Delta V = 6.0 kV = 6000 V is the potential difference between point P and infinity

Substituting into the equation, we find

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

4 0
2 years ago
A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic me
never [62]

1) Current in the wire: 0.0875 A

The current in the wire is given by:

I=\frac{Q}{t}

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

I=\frac{420 C}{4800 s}=0.0875 A

2) Drift velocity of the electrons: 1.78\cdot 10^{-6} m/s

The drift velocity of the electrons in the wire is given by:

u = \frac{I}{nAq}

where

I = 0.0875 A is the current

n=5.8\cdot 10^{28} is the number of free electrons per cubic meter

A is the cross-sectional area

q=1.6\cdot 10^{-19} C is the charge of one electron

The radius of the wire is

r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m

So the cross-sectional area is

A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2

So, the drift velocity is

u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s

4 0
2 years ago
Which of the following is NOT a good way to reduce fuel consumption?
Sidana [21]

Answer:

Explanation:

d

4 0
2 years ago
Read 2 more answers
On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the
Sphinxa [80]

3 trams must be added

Explanation:

In this problem, there are 12 trams along the ring road, spaced at regular intervals.

Calling L the length of the ring road, this means that the space between two consecutive trams is

d=\frac{L}{12} (1)

In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become

d'=(1-\frac{1}{5})d=\frac{4}{5}d

And the number of trams will become

12+n

So eq.(1) will become

\frac{4}{5}d=\frac{L}{n+12} (2)

And substituting eq.(1) into eq.(2), we find:

\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3

Learn more about distance and speed:

brainly.com/question/8893949

#LearnwithBrainly

4 0
2 years ago
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