Answer:
The velocity of the other fragment immediately following the explosion is v .
Explanation:
Given :
Mass of original shell , m .
Velocity of shell , + v .
Now , the particle explodes into two half parts , i.e 
.
Since , no eternal force is applied in the particle .
Therefore , its momentum will be conserved .
So , Final momentum = Initial momentum
The velocity of the other fragment immediately following the explosion is v .
Answer:
90 J
Explanation:
W=fd
W=(75)(1.2)
W= 90 J
Answer:
If efficiency is .22 then W = .22 * Q where Q is the heat input
Heat Input Q = 2510 / .22 = 11,400 J
Heat rejected = 11.400 - 2510 = 8900 J of heat wasted
Also, 8900 J / (4.19 J / cal) = 2120 cal
Explanation:
PE = mgz = 200 * 9.81 *1000 = 1962 KJ
Answer:
281 K
Explanation:
Charles's Law. V1/T1 = V2/T2.
The temperature must be in K = 21.6°C + 273 = 294.6K.
V1T2 = V2T1.
3.62L x T2 = 3.45L x 294.6K
T2 = (3.45 x 294.6) / 3.62 = 1016.4 / 3.62 = B): 281K.
(By direct proportion of volume change: (3.45L / 3.62L) x 294.6K = 281K).
