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Anna [14]
3 years ago
11

Which of the following utilize electromagnetic induction to operate?

Physics
2 answers:
Minchanka [31]3 years ago
7 0

The correct answer is option A and F, Generators in power plants and Solar ovens

Electromagnetic induction is the phenomena of generating electricity due to a changing magnetic field. In electric generators, a rotating coil (which acts as a conductor) is placed between the two magnets. As the coil rotates it induces electromotive force within the conductor. Like wise in case of solar oven , heating is caused due to the eddy currents produced with in the conductor placed with in a changing magnetic field.

kolbaska11 [484]3 years ago
5 0
The correct answer is A. B. C. Hope this helps!!

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2 objects have a total momentum of 400kg m/s, they collide. Object A’s mass is5kg & object B’s mass is 11kg. After the colli
ss7ja [257]

Answer:

Explanation:

We shall apply law of conservation of momentum .

Momentum before collision = momentum after collision .

Momentum before collision = 400 kg m/s

Momentum after collision = 5  x v + 11 x 15

where v is velocity of A after the collision .

5  x v + 11 x 15 = 400

5 v = 400 - 165

5v = 235

v = 47 m /s .

3 0
3 years ago
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vovikov84 [41]

Answer:

D.

boulder-sized rocks that come from the asteroid belt

3 0
3 years ago
1000 kg of water initially at 6 m/s runs through a hydro-generator. If the water leaves the generator at velocity of 4 m/s, and
r-ruslan [8.4K]

Answer:

8.00 kJ

Explanation:

The first thing is to determine what quantities are there.

the mass of water = 1 000 kg

initial velocity, u = 6 m/s

final velocity, v = 4 m/s

the generator is operating at 100 % efficiency, so there is no energy loss.

The kinetic energy, Ek is converted to electrical energy, therefore Ek = electrical energy.

The kinetic energy is calculated as follows:

Ek = 1/2 mv²

    = 1/2×(1 000)× (4)²

    = 8 000 J/s

    = 8.00 kJ  Ans

4 0
3 years ago
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an
Dominik [7]

Answer:

position as a function of time is y = 0.05 × cos(9.9)t

Explanation:

given data

mass = 5 kg

length = 10 cm = 0.1 m

displaced = 5 cm

to find out

position as a function of time

solution

we will apply here equilibrium that is

mass × g = k × length

put here value and find k

k = \frac{5*9.8}{.01}

k = 490 N/m

and ω is

ω = \sqrt{\frac{k}{m} }

ω = \sqrt{\frac{490}{5} }

ω = 9.9

so here position w.r.t  time is

y = 0.05 × cosωt

y = 0.05 × cos(9.9)t

so position as a function of time is y = 0.05 × cos(9.9)t

8 0
3 years ago
A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto
Fed [463]

Answer:

a)   a = 34.375 m / s²,  b)    v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

             v_f = \frac{x-x_1}{t}

we substitute the values

             v_f = \frac{ 6600 -x_1}{4}  

The initial part of the movement is carried out with acceleration

             v_f = v₀ + a t

             x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

             x₁ = ½ a t²

             v_f = a t

we substitute the values

              x₁ = 1/2  a 16²

              x₁ = 128 a

              v_f = 16 a

let's write our system of equations  

               v_f = \frac{6600 - x_1}{4}

               x₁ = 128 a

               v_f = 16 a

we substitute in the first equation  

               16 a = \frac{6600 -128 a}{4}

               16 4 a = 6600 - 128 a

                a (64 + 128) = 6600

                a = 6600/192

                 a = 34.375 m / s²

b) let's find the time to reach this height

                x = ½ to t²

                t² = 2y / a

                t² = 2 5100 / 34.375

                t² = 296.72

                t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

               v_f = 16 a

               v_f = 16 34.375

               v_f = 550 m / s

8 0
3 years ago
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