In real machines, efficiency is always <em>less </em>than 100% .
It still has a definite temperature.
Answer:
Given mass = 2kg, height = 1.2m,g = 9.8.
We know that Work done W = FD
= > W = (mg)(D)
= > W = (2 * 9.8)(1.2)
= > W = 23.52 Joules.
Answer:
1248 ft/ sec
Explanation:
Let Q be the position of the observer and let A be the launch point.
Let B = B(t) be the position of the rocket at time t
Also; Let y = y (t)
Let QA be the height of the rocket at time t
We are to find the at a certain time(t) when the distance from Q to the observer is 13000
If we assume an imaginary right-angle triangle QAB ,
Let h be the length of the hypotenuse QB, By using pythagoras Theorem; we have :
h² = 12000² + y²
Differentiating both sides with respect to t ; we have:
----- Equation (1)
where h = 13000
= 480
Replacing it into above equation to find ; we have:
13000 × 480 = 5000 ×
=
= 1248 ft/sec
∴ The rocket is rising fast at the rate of 1248 feet per second.