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2 answers:
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The rock strike the water with the speed of 15.78 m/sec.
The speed by which rock hit the water is calculated by the formula
v=
v=
v=15.78 m/sec
Hence, the rock strike the water with the speed of 15.78 m/sec.
Answer:
question5: F=74312.5N
question6: charge at the end of antenna=0.37N
Explanation:
Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.
⇒
∴
where is the force of attraction or repulsion
is Coulumb's constant=
and
are the magnitude of the charges
is the distance between two charges
The force between the two charges is attractive if they are of different polarity
The force between the two charges is repulsive if they are of same polarity
Question5:
Given: q1=0.041 C, q2=0.029 C, r=12 m
therefore by Coulumb's law,
Question6:
Given: q1=, r=5 m, F=
therefore by Coulumb's law,
⇒
Answer:
Explanation:
We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:
where me(1) is mechanical energy while on h=10m
and me(2) is mechanical energy while on the ground
Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)
Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.
DynamicE(2) is equal to zero since it's touching the ground
Using that info we have
we divide both sides of the equation with mass to make the math easier.