Answer:
a) t = 0.74s
b) D = 4.76m
c) Vf = 5.35m/s
Explanation:
The ball starts rolling when Vf = ωf*R.
We know that:
Vf = Vo - a*t
ωf = ωo + α*t
With a sum of forces on the ball:
With a sum of torque on the ball:
Replacing both accelerations:
t=0.74s
The distance will be:
Final velocity:
Vf=5.35m/s
Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for DAC:
⇒
Now for the BAC:
⇒
Now, differentiating w.r.t x:
For maximum angle, = 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 =
After solving the above eqn, we get
x =
The observer should stand at a distance equal to x =
Answer:
1. As rocket mass increases, acceleration decreases.
2. The inverse of the mass of the boat.
Explanation:
1. Newton's second law of motion states;
F = ma
where F is the force applied, m is the mass and a is the acceleration.
Therefore, increasing the mass of a rocket increases its weight which would reduce its acceleration provided that the force is constant. Thus, as rocket mass increases, acceleration decreases.
2. The slope of the graph can be expressed as;
From Newton's second law,
F = ma
Slope = (Δa) ÷ (ΔF)
Slope =
⇒ =
Therefore, the slope of the graph is the reciprocal of the mass of the boat.