<span>
In layman's term: </span>like charges don't attract while opposite charges do<span>electrostatic forces between point A( which is charged) and point B (which is also charged) are proportional to the charge of point A and point B. </span><span>there is also something else about this law that I don't quite remember.</span>
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<span />Here is the formula:
<span>F = k x Q1 x Q2/d^<span>2</span></span>
<span>What the formula means:</span>
F=force between charges
Q1 and Q2= amount of charge
d=distance between these two charges
k= Coulombs constant (proportionally constant)
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I think that about covers it and hopefully this helped.
a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J
b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J
Applying Law of Energy conservation :
K 1+U 1
=K 2+U 2
⇒K 1− r 1GmM
=K 2− r 2 GmM
where M=5.0×10 23kg,r1
=> R=3.0×10 6m and m=10kg
(a) If K 1
=5.0×10 7J and r 2
=4.0×10 6 m, then the above equation leads to
K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J
(b) In this case, we require K 2
=0 and r2
=8.0×10 6m, and solve for K 1:K 1
=K 2 +GmM (r 11− r 21)=6.9×10 7 J
Learn more about Kinetic energy on:
brainly.com/question/12337396
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Answer:
For the first one c is the answer
For the second one c is also the answer
For the third one is b
Explanation:
I took that
If you increase the mass m of the car, the force F will increase, while acceleration a is kept constant. Because F and m are directly proportional.
If you increase the acceleration a of the car, the force F will increase, while mass m is kept constant. Because F and a are directly proportional.
How can Newton's laws be verified experimentally; is by setting this experiment, and changing one variable while keeping the other constant, then observe the change in F.
Hope this helps.
Answer:
Explanation:
Theoretical efficiency = T₁ - T₂ / T₁ where T₁ and T₂ is absolute temperature of hot and cold end of the heat engine.
= 600 / (273 + 700 )
= 600 / 973
= .6166
operating efficiency = 40% of .6166
= .4 x .6166
= .2466 = 24.66 %
efficiency = work output / heat input
= 5000 / heat input = .2466
heat input = 5000 / .2466
= 20275.75 J .
HEAT EXTRACED = 20275.75 J.
