What is the maximum number of 7.00 μF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored within the capacitor array under 953 μC ?

Nearly 45 capacitors are required.

Explanation:

It is known that voltage flow in any parallel circuit is tend to be constant. And here the circuit is made up of paralleled arranged capacitors. A capacitor is the one which generates current for a given voltage in the circuit. So the mathematical representation of capacitance is

C = I/V, here I is the current and V is the voltage.

So if  capacitors are connected in parallel, then the sum of varying current will give the equivalent current in the circuit.

I = I1+I2+....In

So,

Since, in the present case, the capacitance is kept constant for all the capacitors connected in parallel and the voltage is constant in parallel circuit connection, then

I= nCV

n = I/CV=953/(3×7) = 45

So nearly 45 capacitors are required.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s