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agasfer [191]
3 years ago
8

A generator contains a coil that spins to produce an electric current. How is energy transformed by using a generator

Physics
1 answer:
LiRa [457]3 years ago
7 0

<u>Answer </u>

Mechanical energy ⇒ Electric energy

<u>Explanation </u>

A generator is used to produce electric energy. When he coils are made to rotate/spin, is makes the mechanical part. This coils rotates inside a magnetic fields inducing current in the coils. When current starts flowing in the coils, it forms the electrical part.

So, on the generator the energy changes are from mechanical to electrical energy.



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A spring of force constant 285.0 N/m and unstretched length 0.230 m is stretched by two forces, pulling in opposite directions a
Kaylis [27]

Answer: W = 0.3853 J, e = 0.052 m

Explanation: Given that,

K =285.0N/M , L = 0.230m , F = 15N , e = ?

F = Ke

15 = 285 × e

e = 15÷ 285

e =0.052 m

e + L = 0.052 + 0.230

= 0.282m ( spring new length )

Work needed to stretch the spring

W = 1/2ke2

W = 1/2 × 285 x 0.052 × 0.052

W = 0.3853 J

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3 years ago
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Explanation:

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2 years ago
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A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin
tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b)  Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

       sin 40 = voy / v₀

       v_{oy} = v₀ sin 40

       v_{oy} = 50.0 sin40

       v_{oy} = 32.14 m / s

       cos 40 = v₀ₓ / V₀

       v₀ₓ = v₀ cos 40

       v₀ₓ = 50.0 cos 40

       v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and v_{y} = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

     p₀ₓ = m v₀ₓ

Y Axis y

    p_{oy} = 0

After the break

X axis

   p_{fx} = m₂ v₂

Axis y

     p_{fy} = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis  

     0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

     v₃ = - m₁ v₁ / m₃

     v₃ = - (-10) 1 / 0.3

     v₃ = 33.3 m / s

X axis

     M v₀ₓ = m₂ v₂

     v₂ = v₀ₓ M / m₂

     v₂ = 38.3  2 / 0.7

     v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

       v_{fy}² = v_{oy}² - 2 gy

       0 =  v_{oy}²-2gy

       y =  v_{oy}² / 2g

       y = 32.14² / 2 9.8

       y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

      v_{f}² =  v_{y}² - 2 g y₂

      0 =  v_{y}² - 2 g y₂

     y₂ =  v_{y}² / 2g

     y₂ = 33.3²/2 9.8

     y₂ = 56.58 m

Total body height is

      Y = y + y₂

      Y = 52.7 + 56.58

     Y = 109.3 m

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3 years ago
What is meant by specific latent heat of fussion​
Lunna [17]

Answer:

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2 years ago
Investigators are researching the appearance of a strange movement of objects in what the local community considers a haunted ho
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Answer:

0.18216 T

Explanation:

N = Number of turns = 219

A = Area = \pi r^2

r = Radius = 1 cm

\omega = Angular speed = 40\times 2\pi

Maximum emf is given by

\epsilon=NBA\omega\\\Rightarrow B=\dfrac{\epsilon}{NA\omega}\\\Rightarrow B=\dfrac{3.15}{219\times \pi 0.01^2\times 40\times 2\pi}\\\Rightarrow B=0.18216\ T

The strength of the magnetic field is 0.18216 T

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