In our lab experiment on ohm's law, the power supply is connected to a circuit containing one resistor, and a direct current of 1.5 a is drawn. if you connect a second identical resistor to the power supply in parallel with the first, (similar to what we did in the lab) how much current is drawn from the battery?

(a) 3.0 a
(b) 0.75 a
(c) 1.5 a
(d) 6.0 a

(A) 3.0=A

Explanation: In order to explain this problem we have to use the OHM law, given by: V=R*I

Besides, we have to consider the resitance equivalent for a parallel connection. This is given by:

1/Re=1/R1+1/R2

If we connect the same resistance, the equivalent resistance is R/2.

Initlally  the current is 1.5 A when one resistance is connected to the batttery. When a second resistance with the same value is connected in parallel to the battery, we have V=Re*Ifinal= (R/2)*Ifinal

also we know that V=R*Iinitial so Iinitial=V/R

then Ifinal= 2*V/R=2*Iinitial

answer: he got 23

explanation:

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2