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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.
The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to

So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by

And if we substitute t=2.7 s, we find the distance covered:
There is no difference. 5 is the same as 5.0
If Resistors are in series= The equivalent is the sum.
E.g R1 and R2 in series, R = R1 + R2.
If in Parallel, equivalent is Product/sum.
E.g If R1 and R2 in parallel, R = (R1*R2)/(R1+R2)
1.) 60 is parallel with 40 and both are then in series with 20.
60//40 = (60*40)/(60+40) = 2400/100 = 24
Now the 24 is in series with the 20
R = 24 + 20 = 44 ohms.
2.) 80 is in series with 40 and both are then in parallel with 40.
Solving the series, R = 80 + 40 =120.
Parallel: 120//40 = (120*40)/(120+40) = 4800/160 = 30
Equivalent Resistance = 30 ohms.
3.) 100 is in parallel with 100 and both are then in series with the parallel of 50 and 50.
The 1st parallel = (100*100)/(100+100) = 10000/200 = 50
The 2nd parallel = (50*50)/(50+50) = 2500/100 = 25.
Solving the series = 50 + 25 =75 ohms.
Cheers.
Answer:
23.8 m
Explanation:
The distance travelled by the zebra can be calculated by using the equation:

where
u is the initial velocity
t is the time
a is the acceleration
For this zebra,
u = 0 since it starts from rest
is the acceleration
Substituting t = 5 s, we find the distance travelled by the zebra:
