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prisoha [69]
3 years ago
10

A straight, stiff, horizontal wire of length 46 cm and mass 20 g is connected to a source of emf by light, flexible leads. A mag

netic field of 1.33 T is horizontal and perpendicular to the wire. 1)Find the current necessary to float the wire, that is, find the current for which the magnetic force balances the weight of the wire.
Physics
1 answer:
dimulka [17.4K]3 years ago
3 0

Answer:

0.32 A

Explanation:

Weight of the wire = m × g  where m, mass = (20 g / 1000 g) × 1 kg = 0.02 kg  and g = 9.8 m/s²

Magnetic force = I L B sinα and α is 90°

and weight of the wire = magnetic force

0.02 kg × 9.8 m/s² = I × 0.46 m × 1.33 × 1

I, current = 0.02 kg × 9.8 m/s² / ( 0.46 m × 1.33T) = 0.32 A

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Which of the following takes place in the combustion chamber of a gas turbine engine?A. A glow plug is used to add enough heat t
erica [24]

Answer:

Fuel oil is mixed with a proper portion of compressed air

Explanation:

A gas turbine has three main part, which are

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The combustion chamber is responsible for mixing fuel with a proper portion of compressed air.

The air compressor supplies air in sufficient quantity to satisfy the requirements of the combustion chamber

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A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper
Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

     ∑  F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

          cos 41 = T₁ₓ / T1

          sin 41 = T₁y / T1

          T₁ₓ = T₁  cos 41

          T₁y= T₁  sin 41

for cable gold

           cos 63 = T₂ / T₂

           sin 63 = T_{2y} / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

              T₃ - W = 0

              T₃ = W

              T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

   - T₁ₓ + T₂ₓ = 0

  T₁ₓ = T₂ₓ

   T1 cos 41 = T2 cos 63

   T1 = T2 cos 63 / cos 41                (1)

y Axis

      T_{1y} + T_{2y} - T3 = 0

       T₁ sin 41 + T₂ sin 63 = T₃          (2)

to solve the system we substitute equation 1 in 2

        T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

         T₂ (cos 63 tan 41 + sin 63) = W

         T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

          T₂ = 200 / (cos 63 tan 41 + sin 63)

          T₂ = 200 / 1,2856

           T₂ = 155.6 N

we substitute in 1

            T₁ = T₂ cos 63 / cos 41

             T₁ = 155.6 cos63 / cos 41

             T₁ = 93.6 N

therefore the tension in each cable is

            T₁ = 93.6 N

             T₂ = 155.6 N

             T₃ = 200 N

6 0
2 years ago
A circuit contains a single 270-pf capacitor hooked across a battery. it is desired to store four times as much energy in a comb
yan [13]
The energy stored by a system of capacitors is given by
U= \frac{1}{2}C_{eq} V^2
where Ceq is the equivalent capacitance of the system, and V is the voltage applied.

In the formula, we can see there is a direct proportionality between U and C. This means that if we want to increase the energy stored by 4 times, we have to increase C by 4 times, if we keep the same voltage.

Calling C_1 = 270 pF the capacitance of the original capacitor, we can solve the problem by asking that, adding a new capacitor with C_x, the new equivalent capacitance of the system C_{eq} must be equal to 4C_1. If we add the new capacitance X in parallel, the equivalent capacitance of the new system is the sum of the two capacitance
C_{eq} = C_1 + C_x
and since Ceq must be equal to 4 C1, we can write
C_1+C_x = 4C_1
from which we find
C_x=3C_1=3 \cdot 270 pF=810 pF
5 0
3 years ago
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