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DiKsa [7]
3 years ago
12

A horse shoe magnet is placed on a mass balance such that a uniform magnetic field of magnitude B runs between it from North to

South. A coil of resistance R is connected to a battery which supplies a potential difference of V across the coil and is suspended such that a section of the coil of length L meters lies between it with current running from East to West. The mass balance measures a mass of M. • What is the measured change in mass due to the effect of Fmag? • What is the total measured mass of the magnets? Keep in mind the effect of Newton's third law.
Physics
1 answer:
BARSIC [14]3 years ago
4 0

Answer:

(a) Measured change in mass (Δm) = BVL/Rg

(b) Total measured mass M' = M - BVL/Rg

Explanation:

Current (I) across is coil is given by the formula;

I = V/R ------------------------1

The magnetic force is given by the formula;

Fb = B*I*L -------------------2

Putting equation 1 into equation 2, we have;

Fb = B*V*L/R -------------------3

Change in mass (Δm) is given as:

Δm = Fb/g -----------------------4

Putting equation 3 into equation 4, we have;

Δm = BVL/Rg

   Therefore,  change in mass (Δm) = BVL/Rg

2. Since B runs from North to South and current running from East to West, then the magnetic force is directed upward.

Therefore,

Total measure mass M' = M - BVL/Rg

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Answer:

A

Since the car is moving to the right, the Normal Force is balancing the Weight and the Net Force is 10 N, right.

Explanation:

as the answers says, the only two forces in the y axis are the normal force and the weight, and they balance each other. On the x axis, you have 20N to the right and the friction is a force that opposes the movement, so the 10N are to the left. The net force is 20 - 10 = 10N.

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3 years ago
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makvit [3.9K]
Its 21!! sorry i was late! :)
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3 years ago
A rotating wheel requires 3.05-s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.05-s interval is 97.9
Setler79 [48]

Answer:

\alpha=14.2rad/s^2

Explanation:

The formula that relates angular displacement with angular acceleration is:

\Delta \theta=\omega_i t+\frac{\alpha t^2}{2}

We can obtain \omega_i from the definition of angular acceleration:

\alpha=\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{t}

\omega_i=\omega_f-\alpha t

Putting all together:

\Delta \theta=(\omega_f-\alpha t) t+\frac{\alpha t^2}{2}=\omega_f t-\frac{\alpha t^2}{2}

Which, since we want the angular acceleration, is:

\alpha=\frac{2(\omega_f t-\Delta \theta)}{t^2}

And for our values is:

\alpha=\frac{2((97.9rad/s)(3.05s)-(37(2\pi rad)))}{(3.05s)^2}=14.2rad/s^2

5 0
3 years ago
Explain why two different elements cannot occupy the same box in the periodic table
bezimeni [28]
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At time t1 = 14 s, a car is located at 99, 80, 27 m and has velocity 4, 0, −3 m/s. At time t2 = 18 s, what is the position of th
Korvikt [17]

Answer:

115, 80, 15m

Explanation

t1 = 14s

t2 = 18s

change in time = 4s (18-14)

r(final) = r(initial) + (average velocity) x (change in time)

multiply the average velocity with the change in time

= (4, 0, -3) x 4 = 16, 0, -12

now we'll add this value to the initial position of the car

(99, 80, 27)m + (16, 0, -12)m = (115, 80, 15)m

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4 years ago
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