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IgorLugansk [536]
3 years ago
6

A closely wound search coil has an area of , turns, and a resistance of . It is connected to a

Physics
1 answer:
Pepsi [2]3 years ago
8 0

Answer:

The magnitude of  field is 0.095T.

Explanation:

Please see attachment .

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HELP URGENT!!!!!!!!!!!!!!!!!!!!!!!!
nirvana33 [79]
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Because the rest of the answers are illogical
5 0
3 years ago
When you set a heavy bag down on the ground, you are doing _______ work on it.
puteri [66]

When you set a heavy bag down on the ground, you are doing negative work on it.

4 0
2 years ago
To complete a project, 200,000 Joules of work is needed. The time taken to complete the project is 20 seconds. How much power is
Leno4ka [110]

Answer:

10,000

Explanation:

200,000/20 = power needed

200,000/20 = 10,000

Hope this Helps!

5 0
3 years ago
Read 2 more answers
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
3 years ago
This is the change in kinetic energy of a system in which a 16 kg object moving at 25 m/s slows to a velocity of 20 m/s
Dennis_Churaev [7]

The kinetic energy of an object is given by

KE = 0.5mv²

where m is the mass and v is the velocity.

To calculate the change in kinetic energy...

Initial KE:

KEi = 0.5mVi²

where Vi is the initial velocity.

Final KE:

KEf = 0.5mVf²

where Vf is the final velocity.

ΔKE = KEf - KEi

ΔKE = 0.5mVi² - 0.5mVf²

ΔKE = 0.5m(Vf²-Vi²)

Given values:

m = 16kg

Vi = 25m/s

Vf = 20m/s

Plug in the given values and solve for ΔKE:

ΔKE = 0.5×16×(20²-25²)

ΔKE = -1800J

5 0
3 years ago
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