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mr Goodwill [35]
3 years ago
12

5. Suppose that you wish to construct a simple ac generator having an output of 24 V maximum when rotated at 120 Hz. A uniform m

agnetic field of 0.10 T is available. If the area of the rotating coil is 100 cm2, how many turns do you need
Physics
1 answer:
vazorg [7]3 years ago
6 0

Answer:

32 turns

Explanation:

From the expression for the induced emf,

E = (N)(B)(A) w

E = emf = 24 V

N = number of turns = ?

B = magnetic field strength = 0.10 T

A = Cross sectional Area of the loop = 100 cm² = 0.01 m²

w = Angular speed = (2πf) = (2π × 120) = 754.3 rad/s

24 = N (0.1)(0.01)(754.3)

N = (24/0.7543)

N = 31.8 ≈ 32 turns.

Hope this Helps!!!

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An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.60 m above
s344n2d4d5 [400]

Answer:

a) No

b) the rock must have a minimum initial speed of 6.79m/s for it to reach the top of the building.

Explanation:

Given:

Height of the wall = 3.95m

Initial height = 1.60m

Initial speed = 5.00m/s

distance between the initial height and wall top = 3.95 - 1.60 = 2.35m

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v^2 = u^2 + 2as ....1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled

From equation 1

s = (v^2 - u^2)/2a ...2

Since the rock t moving up,

the acceleration = -g = -9.8m/s2

s = maximum height travelled

v = 0 (at maximum height velocity is zero)

Substituting into equation 2

s = (0 - 5^2)/(2×-9.8) = 1.28m

Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall

b) Using equation 1:

u^2 = v^2 - 2as

v = 0

a = -9.8m/s

s = 2.35m. (distance between the initial height and wall top)

u^2 = 0 - 2(-9.8 × 2.35)

u^2 = 46.06

u = √46.06

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Now we know that the row is located 5.6 cm from the centre of the disc:

r = 5.6 cm = 0.056 m

So we can find the tangential speed of the row as the product between the angular speed and the distance of the row from the centre of the circle:

v=\omega r = (125.6 rad/s)(0.056 m)=7.0 m/s

b.  875 m/s^2

The acceleration of the row of data (centripetal acceleration) is given by

a=\frac{v^2}{r}

where we have

v = 7.0 m/s is the tangential speed

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Substituting numbers into the formula, we find

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3 years ago
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