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MaRussiya [10]
3 years ago
15

An isolated conducting sphere has a 15 cm radius. One wire carries a current of 1.0000046 A into it while another wire carries a

current of 1.0000000 A out of it. How long in seconds would it take for the sphere to increase in potential by 890 V?

Physics
1 answer:
lbvjy [14]3 years ago
7 0

Answer: It would take 3.23ms

Explanation: Please see the attachments below

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A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat
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Power is needed for (1) acceleration and (2) lifting the loaded chairs. These two parts can be calculated separately and then added together.

(1) Power for acceleration:

The final speed of the lift is

V=(10 km/h)(1 h×1000 m60 sec×1 km)=2.887 m/s.

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Pa=12m(V2−V20)/Δt=12(50×250 kg)(2.778 m/s)2=9.6 kW.

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Assume that the acceleration is constant (i.e. power supply is constant), its value will be

a=ΔVΔt=2.778 m/s5 s=0.556 m/s2.

Then the vertical lift during acceleration will be

(12at2)×(2001000)=1.36 m.

Hence, the power needed to increase the potential energy of the lift is

Pg=mgΔhΔt=(50×250 kg) (9.89 m/s2)(1.36 m)/(5 s)=3.41 kW.

Then the total Power required is

Ptotal=Pa+Pg=9.6+34.1=43.7 kW.

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Describe what introspection is.
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Answer:

Introspection is a process that involves looking inward to examine one's own thoughts and emotions. ... The experimental use of introspection is similar to what you might do when you analyze your own thoughts and feelings but in a much more structured and rigorous way.

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Show your work and resoning for the below requirement.
Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

 They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

            - T_{y} L + W₂ L + W₁ L / 2 = 0

            T_{y} = W₂ + W₁ / 2

            T_{y} = 120 + 150/2

            T_{y} = 195 lb

we use trigonometry to find the cable tension

             sin 30 = T_{y} / T

             T = T_{y} / sin 30

             T = 195 / sin 30

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This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

             T < 500 lb

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4 years ago
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