A simple pendulum takes 32 sec to complete 20 oscillations .what is the time period of the pendulum?
1 answer:
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Answer:
a) 23.51 m/s
b) 1.07 kg
Explanation:
Parameters given:
Kinetic energy, K = 295 J
Momentum, p = 25.1 kgm/s
a) The kinetic energy of a body is given as:
where m = mass of the body and v = speed of the body
We know that momentum is given as:
p = mv
Therefore:
K = 1/2 * pv
=> v = 2K / p
v = (2 * 295) / 25.1 = 23.51 m/s
The velocity of the body at that instant is 23.51 m/s.
b) Momentum is given as:
p = mv
=> m = p / v
m = 25.1 / 23.51 = 1.07 kg
The mass of the body at that instant is 1.07 kg
Refer to the diagram shown below.
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω