Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
The new natural frequency would be ω/2.
we know that,
= ω. -> equation 1
now, when capacitance is quadrupled,

. -> equation 2
substituting value of equation 1 in equation 2 , we get,

Hence, the new natural frequency of the circuit is ω/2.
what do you mean by frequency ?
The resonant frequency for a particular circuit is the frequency at which this equality stands true. Where L is the inductance in henries and C is the capacitance in farads, this is the LC circuit's resonant frequency.
Learn more about frequency here:-
brainly.com/question/12530980
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Answer:3,45 x 10^9 N
Explanation: We have considered the total charge for each coin , this is the total atoms x 29 electrons for cooper and multiplier by electron charge, the total charge for each coin is 0,464 C
Finally we use the Coulomb law,
F=k Q/ (r)^2
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