The driver is tooling along in his snowmobile, pointed north,
at 8.5 m/s.
He's carrying the flares with him, so the flares are also moving north
at 8.5 m/s.
When he fires the flare straight up, it has a vertical velocity of 4.3 m/s
straight up, and a horizontal velocity of 8.5 m/s towards the north.
The magnitude of the net velocity is √(4.3² + 8.5²) .
That's about 9.53 m/s, at some angle between straight up
and straight north.
The angle above horizontal is the angle that has a tangent of 4.3/8.5 .
I'll let you work out the angle.
Answer:
(a) k = 30.33 N/m
(b) a = 9.8 m/s²
Explanation:
First, we need to find the force acting on the bungee jumper. Since, this is a free fall motion. Therefore, the force must be equal to the weight of jumper:
F = W = mg
F = (65 kg)(9.8 m/s²)
F = 637 N
(a)
Now applying Hooke's Law:
F = k Δx
where,
k = spring constant = ?
Δx = change in length of bungee cord = 33 m - 12 m = 21 m
Therefore,
637 N = k(21 m)
k = 637 N/21 m
<u>k = 30.33 N/m</u>
<u></u>
(b)
Since, this is free fall motion. Thus, the maximum acceleration will be the acceleration due to gravity.
a = g
<u>a = 9.8 m/s²</u>
Answer:
Explanation:
Initial velocity of mailbag u = 2 m/s
acceleration downwards a = g = 9.8 m/s²
time t = 3 s
a ) final velocity v = ?
v = u + at
= 2 + 9.8 x 3
= 31.4 m /s
b )
s = ut + 1/2 g t²
s is relative displacement of mailbag
u = relative initial velocity of mailbag = 0
relative acceleration = g = 9.8 m /s²
time t = 3 s
s = 0 + 1/2 x 9.8 x 3²
= 44.1 m
relative displacement of mailbag = 44.1 m .
