A supersonic airplane is flying horizontally at a speed of 2650 km/h. What is the centripetal acceleration of the airplane, if i
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Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;
Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.
By definition the exchange of heat is given by
where,
m = mass
c = specific heat
= Change in temperature
Therefore the total heat exchange is given as
Our values are given as,
Total mass is = 200lb ,however the mass of solid vegetable and water is given as,
Replacing at our equation we have,
Therefore the heat removed is 22411.2 Btu