Work done is the distance a force acts over.
So, the work done here is 9.0N * 3.0m = 27 J
Answer:
(A) L = 115.3kgm²/s
(B) dL/dt = 94.1kgm²/s²
Explanation:
The magnitude of the angular momentum of the rock is given by the foemula
L = mvrSinθ
We have been given θ = 36.9°, m = 2.0kg, v = 12.0m/s and r = 8.0m.
Therefore L = 2.00 × 12 × 8.0 × Sin 36.9° =
115.3 kgm²/s
(B) The magnitude of the rate of angular change in momentum is given by
dL /dt = d(mvrSinθ)/dt = mgrSinθ = 2.00 × 9.8 × 8.0× Sin36.9 = 94.1kgm²/s²
Let's just assume that you throw the ball with an initial speed of 2 m/s instead of dropping it like free falling.
a=9.81 m/s^2
Vi= 2 m/s
t= 3 x
we use the formula
d = (Vi)(t) + (1/2)(a)(t)^2
d= (2)(3) + (1/2)(9.81)(9)
d=50.145 m
Lets assume that s is the speed of the slower train and f is the speed of the faster train.
s = f - 16
f = 150 / t
s = 170 / ( t+2 )
-----------------------
170 / ( t+2 ) = 150 / t - 16 / t * ( t + 2 )
170 t = 150 * ( t + 2 ) - 16 t * ( t + 2 )
170 t = 150 t + 300 - 16 t² - 32 t
16 f² + 52 t - 300 = 0 / : 4 ( We will divide both sides of the equation by 4 )
4 t² + 13 t - 75 = 0
t 1/2 = ( -13 + √(169 + 1200 ) )/ 8
t = ( - 13 + 37 ) / 8 = 24 / 8 = 3
t = 3 h
s = 170 : ( 3 + 2 ) = 170 : 5 = 34 mph
Answer: The speed of the slower train is 34 mph.
We can answer the question by looking at the Ohm's law, which gives us the relationship between voltage (V), current (I) and resistance (R) of a circuit:

equivalently, we can rewrite it as

by looking at the equation, we can make the following observations:
1) The current is proportional to the voltage: therefore, if the voltage increases, the current increases as well; if the voltage decreases, the current decreases too.
2) The current is inversely proportional to the resistance: if the resistance increases, the current decreases, and if the resistance decreases, the current increases.