Teams red and blue are having a tug-of-war. According to Newton's third law, the force with which the red team pulls on the blue
1 answer:
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Answer:
m = 3.91 kg
Explanation:
Given that,
Mass of the object, m = 3.74 kg
Stretching in the spring, x = 0.0161 m
The frequency of vibration, f = 3.84 Hz
When the object is suspended, the gravitational force is balanced by the spring force as :
k = 2276.52 N/m
The frequency of vibration is given by :
m = 3.91 kg
So, the mass of the object is 3.91 kg. Hence, this is the required solution.
Answer:
a) the required time is 0.6283 μs
b) the inductor current is 0.5 mA
Explanation:
Given the data in the question;
The capacitor voltage has its maximum value of 25 V at t = 0
i.e V = V₀ = 25 V
we determine the angular velocity;
ω = 1 / √( LC )
ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )
ω = 1 / √( 1.6 × 10⁻¹³ )
ω = 1 / 0.0000004
ω = 2.5 × 10⁶ s⁻¹
a) How much time does it take until the capacitor is fully discharged for the first time?
V = V₀sin( ωt )
we substitute
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
divide both sides by 25 V
sin( 2.5 × 10⁶ × t ) = 1
( 2.5 × 10⁶ × t ) = π/2
t = 1.570796 / (2.5 × 10⁶)
t = 0.6283 × 10⁻⁶ s
t = 0.6283 μs
Therefore, the required time is 0.6283 μs
b) What is the inductor current at that time?
(t) = V₀√(C/L) sin(ωt)
{ sin(ωt) = 1 )
(t) = V₀√(C/L)
we substitute
(t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )
(t) = 25 × 0.00002
(t) = 0.0005 A
(t) = 0.5 mA
Therefore, the inductor current is 0.5 mA