A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency
1 answer:
Answer:
Explanation:
f =
T = 120 N
L = 3.00 m
(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m
(wow that's massive for a "rope")
f = )
f = /6 = 0.527 Hz
This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.
A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N
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Answer:
the final temperature of the tea is 7.39⁰C.
Explanation:
Given;
mass of the tea, m = 375 g
specific heat capacity of the tea, C = 4.184 JJ/g°C
initial temperature of the tea, t₁ = 95°C
the final temperature of the tea, t₂ = ?
Energy lost by the refrigerator, Q = 137,460 J
The energy lost by the refrigerator is given by the following formula;
-Q = mc(t₂ - t₁)
-137,460 =375 x 4.184(t₂ - 95°C)
-137,460 = 1569(t₂ - 95°C)
Therefore, the final temperature of the tea is 7.39⁰C.