Because iron is a metal and cobalt is a non-metal
You are exerting 100N. Since there’s no NET force, then there must be exactly 100N pushing exactly back on your 100N to cancel it to exactly zero. Newton's first law states that whether a body is at rest or travelling in a straight line at a constant speed, it will remain at rest or continue to move in a straight line at a constant speed unless acted upon by a force.
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:
f(t) = 28,7 [N]
Explanation: IMPORTANT NOTE: IN PROBLEM STATEMENT CHARGES ARE IN C (COULOMBS) AND IN THE DIAGRAM IN μC. WE ASSUME CHARGES ARE IN μC.
The net force on +q₂ is the sum of the force of +q₁ on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)
The two forces have the same direction to the right of charge q₂, we have to add them
Then
f(t) = f₁₂ + f₃₂
f₁₂ = K * ( q₁*q₂ ) / (0,1)²
q₁ = + 8 μC then q₁ = 8*10⁻⁶ C
q₂ = + 3,5 μC then q₂ = 3,5 *10⁻⁶ C
K = 9*10⁹ [ N*m² /C²]
f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻² [ N*m² /C²]* C*C/m²
f₁₂ = 252*10⁻¹ [N]
f₁₂ = 25,2 [N]
f₃₂ = 9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²
f₃₂ = 78,75*10⁻³/ 2,25*10⁻²
f₃₂ = 35 *10⁻¹
f₃₂ = 3,5 [N]
f(t) = 28,7 [N]
The battery doesn't 'use' power. The battery <em>produces</em> the power that all the other electrical devices use.
If the starter motor is using 2,520 watts, then the battery is producing energy at the rate of 2,520 watts. That means <em>2,520 Joules</em> of energy every second.
Thanks for giving us the formula.
E = P x t
Energy = Power x Time
Energy = (2,520 watts) x (1 second)
Energy = 2,520 Joules
