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Oksana_A [137]
3 years ago
14

A river is flowing south at a rate of 3 m/s. Steven can roe directly across the river if he aims the raft 30 degrees. What rate

is Steven rowing at
Physics
1 answer:
levacccp [35]3 years ago
3 0

Answer:

Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s

Explanation:

Given data,

The river flowing south at the rate, v = 3 m/s

To reach the other side directly across the river, he aims the raft, Ф = 30°

The speed of his raft across the river is given by the formula,

                                          V = v / Sin Ф

                                             = 3 / Sin 30°

                                              = 6 m/s

Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s

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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
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Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

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\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

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Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

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