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Sedaia [141]
3 years ago
15

A 0.40-μF capacitor is connected to a 5.0-V battery. How much charge is on each plate of the capacitor? Express your answer usin

g two significant figures.
Physics
1 answer:
weqwewe [10]3 years ago
7 0

Answer:

Q=2.0\times10^{-6}C

Explanation:

The equation of charge on the capacitor given its capacitance and a voltage applied to it is Q=CV.

Our capacitance is expressed in <em>microfarads</em>, where micro stands for \times10^{-6}, so in S.I. we sill have:

Q=CV=(0.4\times10^{-6}F)(5V)=0.000002C=2.0\times10^{-6}C

Where the results has been expresed with <em>two significant figures</em>.

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In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of t
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Answer:

Velocity of airplane is 500 km/h

Velocity of wind is 40 km/h

Explanation:

V_a= Velocity of airplane in still air

V_w= Velocity of wind

Time taken by plane to travel 1150 km against the wind is 2.5 hours

V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)

Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours

V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)

Subtracting the two equations we get

V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h

Applying the value of velocity of wind to the first equation

V_a-40=460\\\Rightarrow V_a =500\ km/h

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

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