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Agata [3.3K]
3 years ago
8

A rectangular coil with sides 0.130 m by 0.250 m has 513 turns of wire. It is rotated about its long axis in a uniform magnetic

field of 0.550 T that is perpendicular to the long axis. At what frequency, in complete cycles per second, must the coil be rotated for it to generate a maximum voltage of 115 V?
Physics
1 answer:
Sonbull [250]3 years ago
3 0
Voltage = N * Δ(BA)/Δt 
<span>BA = 0.57*0.16*0.22 = 2.0064e-2 </span>
<span>N = 505 </span>
<span>115/505 = Δ(BA)/Δt = 23/101 </span>
<span>When the top of the coil rotates to the bottom (1/2 half cycle) BA changes from max to min and when the bottom rotates back to the top BA changes from min to max. So Δ(BA) is twice per cycle </span>
<span>So 2*101Δ(BA)=23Δt and Δt = 1/f </span>
<span>202*2.0064e-2/23 =Δt = 1/f => f =5.675Hz</span>
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7 0
2 years ago
A crane lifts an air conditioner to the top of a building. If the building is 12 m high, and the air conditioner has a mass of 2
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Work needed = 23,520 J

<h3> Further explanation </h3>

Given

height = 12 m

mass = 200 kg

Required

work needed by the crane

Solution

Work is the transfer of energy caused by the force acting on a moving object  

Work is the product of force with the displacement of objects.  

Can be formulated  

W = F x d  

W = Work, J, Nm  

F = Force, N  

d = distance, m  

F = m x g

Input the value :

W = mgd

W = 200 kg x 9.8 m/s²x12 m

W = 23520 J

8 0
3 years ago
Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which
antiseptic1488 [7]

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = K_{f} - K₀

          K_{f}  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        K_{f}  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

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5 0
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Andre45 [30]

Answer:

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3 years ago
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pishuonlain [190]
The answer is false
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