Answer:
The time he can wait to pull the cord is 41.3 s
Explanation:
The equation for the height of the skydiver at a time "t" is as follows:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height at time "t".
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
First, let´s calculate how much time will it take for the skydiver to hit the ground if he doesn´t activate the parachute.
When he reaches the ground, the height will be 0 (placing the origin of the frame of reference on the ground). Then:
y = y0 + v0 · t + 1/2 · g · t²
0 m = 15000 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²
0 m = 15000 m - 4.9 m/s² · t²
-15000 m / -4.9 m/s² = t²
t = 55.3 s
Then, if it takes 4.0 s for the parachute to be fully deployed and the parachute has to be fully deployed 10.0 s before reaching the ground, the skydiver has to pull the cord 14.0 s before reaching the ground. Then, the time he can wait before pulling the cord is (55.3 s - 14.0 s) 41.3 s.
Answer:
v = 0.84m/s, v(max)= 0.997m/s
Explanation:
Initial work done by the spring, where c is the compression = 0.28m:
Work lost to friction:
Energy:
(a) Solve for v:
(b) Solve 
for x:
if:
Answer:
the relative speed should be 40mph
Explanation:
Answer:
38,437.5
Explanation:
Density(d)= 102.5g/ml
Volume (v)=375ml
Mass(m) = ?
D =m/v
102.5= m/375
102.5*375=m
38,437.5=m
therefore Mass = 38,437.5g/ml.
