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faltersainse [42]
3 years ago
5

The unit used to measure electric current is the ampere (AA). Now, assume that the current delivered at a wall socket reaches th

e value 3.8 AA ten times in a time interval of 0.17 ss. What is the period TTT with which the current at the wall socket changes?
Physics
1 answer:
Lostsunrise [7]3 years ago
8 0

The period T with which the current at the wall socket changes is 0.017 s.

It also ranges from 0.155 s to 0.0189 s since the cycles is not given in the question.

<u>Explanation</u>:

The current reaches 3.8 A once for every cycle. So there must have been 10 cycles (10 periods) in 0.17 s.  

Using 'T' for period:

                                10 T = 0.17  

                                      T = 0.17 / 10  

                                     T  = 0.017 s.

The question doesn't tell about the cycles and so we assume it as 10 cycles. If there are 9 cycles due to the current 3.8 A at the exact beginning and end of 0.17 s. This would make the period T as

                              0.17 / 9 = 0.0189 s approx.

Likewise, if the 3.8 A occurs at the end of the 1st cycle, 11 cycles would have passed during 0.17 s.

In that case the period = 0.17 / 11 = 0.155 s approx.

So T could be in the range from 0.155 s to 0.0189 s.

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MrRa [10]

Answer:

Option B. 3.0×10¯¹¹ F.

Explanation:

The following data were obtained from the question:

Potential difference (V) = 100 V.

Charge (Q) = 3.0×10¯⁹ C.

Capacitance (C) =..?

The capacitance, C of a capacitor is simply defined as the ratio of charge, Q on either plates to the potential difference, V between them. Mathematically, it is expressed as:

Capacitance (C) = Charge (Q) / Potential difference (V)

C = Q/V

With the above formula, we can obtain the capacitance of the parallel plate capacitor as follow:

Potential difference (V) = 100 V.

Charge (Q) = 3.0×10¯⁹ C.

Capacitance (C) =..?

C = Q/V

C = 3.0×10¯⁹ / 100

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Therefore, the capacitance of the parallel plate capacitor is 3.0×10¯¹¹ F.

7 0
3 years ago
Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

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         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

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let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

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        v = √65.44

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statuscvo [17]
You can use the impulse momentum theorem and just subtract the two momenta.
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loris [4]

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