It’s a good movie maybe. Since I like it as well
Answer:
w=255
Explanation:
The change in internal energy is given by the first law:
ΔE = Q - w
where ΔE is the change in internal energy of the system
q is the heat added to the system
w is the work done *by* the system on the surroundings
So, for the first phase of this process:
ΔE = Q - w
Q=160J
w=309J
ΔE = 160J - 309J = -149J
To bring the system back to its initial state after this, the internal energy must change by +149J (the system myst gain back the 149 J of energy it lost). We are told that the system loses 106 J of heat in returning to its initial state, so the work involved is given by:
ΔE = Q - w
+149J = -106J - w
255J = -w
w = -255J
Explanation:
(a)
The initial vertical velocity is 13 m/s. At the maximum height, the vertical velocity is 0 m/s.
v = at + v₀
0 = (-9.8) t + 13
t ≈ 1.33 s
(b)
Immediately prior to the explosion, the ball is at the maximum height. Here, the vertical velocity is 0 m/s, and the horizontal velocity is constant at 25 m/s.
v = √(vx² + vy²)
v = √(25² + 0²)
v = 25 m/s
(c)
Momentum is conserved before and after the explosion.
In the x direction:
m vx = ma vax + mb vbx
m (25) = (⅓ m) (0) + (⅔ m) (vbx)
25m = (⅔ m) (vbx)
25 = ⅔ vbx
vbx = 37.5 m/s
And in the y direction:
m vy = ma vay + mb vby
m (0) = (⅓ m) (0) + (⅔ m) (vby)
0 = (⅔ m) (vby)
vby = 0 m/s
Since the vertical velocity hasn't changed, and since Fragment B lands at the same height it was launched from, it will have a vertical velocity equal in magnitude and opposite in direction as its initial velocity.
vy = -13 m/s
And the horizontal velocity will stay constant.
vx = 37.5 m/s
The velocity vector is (37.5 i - 13 j) m/s. The magnitude is:
v = √(vx² + vy²)
v = √(37.5² + (-13)²)
v ≈ 39.7 m/s
By looking at how wiggily the bar is lol
