The main formula to be used here is
Force = (mass) x (acceleration).
We'll get to work in just a second. But first, I must confess to you that I see
two things happening here, and I only know how to handle one of them. So
my answer will be incomplete, but I believe it will be more reliable than the
first answer that was previously offered here.
On the <u>right</u> side ... where the 2 kg and the 3 kg are hanging over the same
pulley, those weights are not balanced, so the 3 kg will pull the 2kg down, with
some acceleration. I don't know what to do with that, because . . .
At the <em>same time</em>, both of those will be pulled <u>up</u> by the 10 kg on the other side
of the upper pulley.
I think I can handle the 10 kg, and work out the acceleration that IT has.
Let's look at only the forces on the 10 kg:
-- The force of gravity is pulling it down, with the whatever the weight of 10 kg is.
-- At the same time, the rope is pulling it UP, with whatever the weight of 5 kg is ...
that's the weight of the two smaller blocks on the other end of the rope.
So, the net force on the 10 kg is the weight of (10 - 5) = 5 kg, downward.
The weight of 5 kg is (mass) x (gravity) = (5 x 9.8) = 49 newtons.
The acceleration of 10 kg, with 49 newtons of force on it, is
Acceleration = (force) / (mass) = 49/10 = <em>4.9 meters per second²</em>
Answer:
The equivalent resistance of the parallel circuit would be 20 Ω
Explanation:
To calculate the resistance of resistors connected in parallel, the formula to be used is
1/R = 1/R₁ + R₂ + R₃ + R₄...
1/R = 1/120 + 1/60 + 1/40
1/R = (1 + 2 + 3)/120
1/R = 6/120
1/R = 1/20 Ω
This can be rewritten or cross-multiplied to be
R × 1 = 20 × 1
R = 20 Ω
The equivalent resistance (R) would then be 20 Ω
