Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w =
=
= 18.38 rad/sec
Period of oscillation = 
= 0.34 sec
All you would do is for a, 10 times 2 is 20 so it would be 20-dB
For b, 10 times 4 is 40 so it would be 40-dB
<span>For c, 10 times 8 is 80 so it would be 80-dB</span>
Answer:
is 3 and 2
Explanation: the firth one is 3 and the 2
Magnitude of displacement = 
Adding the squares gives displacement = 
Displacement =
≈ 724.7m
Explanation:
Amperage is the unit of electric current. It describes the strength of the electric current in a circuit.
The voltage is the driving force of the current in a circuit
Power is a function of voltage and current in the circuit.
Current is designate as I
Voltage as V
Power as P
I = 
Where R is the resistance to flow of electricity
P = I x V = 
The unit of power is watts and voltage is volts
learn more:
Voltage brainly.com/question/6949231
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