Consider the following. (Let C1 = 20.80 µF and C2 = 14.80 µF.) A rectangular circuit contains a battery and four capacitors. The
bottom side has a 9.00 V battery with the positive terminal on the left. The left and right sides of the circuit each contain a capacitor labeled C1. The top side splits into two parallel horizontal branches, which recombine before reaching the top right corner. There is a 6.00 µF capacitor on the upper branch and a capacitor labeled C2 on the lower branch.
(a) Find the equivalent capacitance of the capacitors in the figure. µF
(b) Find the charge on each capacitor. on the right 20.80 µF capacitor µC on the left 20.80 µF capacitor µC on the 14.80 µF capacitor µC on the 6.00 µF capacitor µC
(c) Find the potential difference across each capacitor. on the right 20.80 µF capacitor V on the left 20.80 µF capacitor V on the 14.80 µF capacitor V on the 6.00 µF capacitor V
(a) Find the equivalent capacitance of the capacitors in the figure. µF
(b) Find the charge on each capacitor. on the right 20.80 µF capacitor µC on the left 20.80 µF capacitor µC on the 14.80 µF capacitor µC on the 6.00 µF capacitor µC
(c) Find the potential difference across each capacitor. on the right 20.80 µF capacitor V on the left 20.80 µF capacitor V on the 14.80 µF capacitor V on the 6.00 µF capacitor V
1 answer:
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The tennis ball lands at a point 40.4 m from the base of the building.
The tennis ball is projected with a horizontal velocity <em>u</em> from a window, which is at a height <em>y</em> from the ground. The ball lands at a distance <em>x</em> from the base of the building. Let the ball take a time <em>t</em> to reach the ground. In the time <em>t</em> ,the ball falls a vertical distance <em>y</em> and also travel a horizontal distance <em>x</em>.
The initial vertical velocity of the ball is zero, since the ball is projected in the horizontal direction. The ball falls down under the action of gravitational force.
Thus, use the equation of motion,
rewrite the expression for <em>t</em> and calculate the value of <em>t</em> using 9.81 m/s²for <em>g</em> and 500 m for <em>y</em>.
The horizontal distance <em>x</em> is traveled using the constant velocity <em>u </em>since no force acts on the ball in the horizontal direction.
Therefore,
Substitute 4 m/s for <em>u</em> and 10.096 s for <em>t</em>
Thus, the ball lands at a point 40.4 m from the base of the building.
Explanation:
Given data
Pressure P=1200 kPa
To find
Pressure in
(a) kN/m²
(b) kg/m.s²
(c) kg/km.s²
Solution
For Part (a)
For Part (b)
For Part (c)