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asambeis [7]
3 years ago
9

If a coil is rotated in a magnetic field at twice the speed -- ie twice the frequency -- the amplitude of the EMF induced in the

loop will...
A. remain unchaged

B. reduce by half

C. double
Physics
1 answer:
valkas [14]3 years ago
6 0

Answer:

C. double

Explanation:

The voltage of a EMF depends largely on the amount of spins that the coild is able to do, in this case as yuo are doubling the speed at which the coil is rotating you are also doubling the number of spins that the coil is doing in a certain amount of time, so the amplitude of the EMF induce on the loop will also double.

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A car weighing 12,000 N is parked on a 36° slope. The car starts to roll down the hill. What is the acceleration of the car?​
Delicious77 [7]

Answer: 5.8 m/s squared

Explanation: just got that question lol

4 0
2 years ago
If a boat and its riders have a mass of 900 kg and the boat drifts in at 1.4 m/s how much work does sam do to stop it?
Rama09 [41]
The initial kinetic energy of the boat and its rider is
K_i =  \frac{1}{2} mv_i^2 =  \frac{1}{2}(900 kg)(1.4 m/s)^2=882 J

After Sam stops it, the final kinetic energy of the boat+rider is
K_f = 0 J
because its final velocity is zero.

For the law of conservation of energy, the work done by Sam is the variation of kinetic energy of the system:
W=K_f-K_i =0-882 J=-882 J
where the negative sign is due to the fact that the force Sam is applying goes against the direction of motion of the boat.
6 0
3 years ago
In an electrostatic field, path 1 between points A and B is twice as long as path 2. The electrostatic work done on a negatively
Elanso [62]

Answer:

W2 = W1

Explanation:

work is independent of the path taken between the points.

8 0
2 years ago
A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.
Stella [2.4K]

Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

\omega =\frac{2\pi \times 469}{60}=49.11 rad/s

t=16 s

Angular deceleration in 16 s

\omega =\omega _0+\alpha \cdot t

\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2

Moment of Inertia I=mr^2=13\times 1.8^2=42.12 kg-m^2

Change in kinetic energy =Work done

Change in kinetic Energy=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}

\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J

(a)Work done =50.79 kJ

(b)Average Power

P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW

7 0
3 years ago
Pls help me with this problem!!
Olenka [21]

Answer:

v = 19.6 m/s.

Explanation:

Given that,

The radius of the circle, r = 5 m

The time period of the ball, T = 1.6s

We need to find the ball's tangential velocity.

The formula for the tangential velocity is given by :

v=\dfrac{2\pi r}{T}

Putting all the values in the above formula

v=\dfrac{2\pi \times 5}{1.6}\\\\v=19.6\ m/s

So, the tangential velocity of the ball is 19.6 m/s. Hence, the correct option is (c).

7 0
3 years ago
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