Answer:
(a) Vf = 128 ft/s
(b) K.E = 122.8 Btu
Explanation:
(a)
In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = 32.2 ft/s²
h = height = 253 ft
Vf = Final Velocity = ?
Vi = Initial Velocity = 10 ft/s
Therefore,
(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²
16293.2 ft²/s² + 100 ft²/s² = Vf²
Vf = √(16393.2 ft²/s²)
<u>Vf = 128 ft/s</u>
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(b)
The kinetic energy of the object before it hits the surface of earth is given by:
K.E = (0.5)(m)(Vf)²
where,
m = mass of object = 375 lb
K.E = Kinetic energy of object before it strikes the surface of earth = ?
Therefore,
K.E = (0.5)(375 lb)(128 ft/s)²
K.E = 3073725 lb.ft²/s²
Now, converting this to Btu:
K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)
<u>K.E = 122.8 Btu</u>
Answer:
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Answer:
0.12m/s
Explanation:
v=λf
Given that, λ = 12cm = 0.12m
T = 1second
(A period T is the time required for one complete cycle of vibration to pass a given point)
frequency 'f' is unknown but we can get frequency from f = 1/T = 1/1 = 1Hz
therefore, v= 0.12 × 1 = 0.12m/s
