A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is i
1 answer:
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
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Explanation:
Expression to calculate thermal resistance for iron () is as follows.
where, = length of the iron bar
= thermal conductivity of iron
= Area of cross-section for the iron bar
Thermal resistance for copper () = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]
where, = length of copper bar
= thermal conductivity of copper
= Area of cross-section for the copper bar
Now, expression for the transfer of heat per unit cell is as follows.
Q =
Putting the given values into the above formula as follows.
Q =
=
= 2.92 Joule
It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,
P =
Here, T is 1 second so, power conducted is equal to heat transferred.
So, P = 2.92 watt
Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.