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: 4
Explanation: because 0s aren’t significant and after the decimal point, there was to be a value greater than 0 than the rest are sig figs.
1.) Pitch
2.)Wavelength
3.)Density/Elastic Properties-b. Two of the above
4.)Liquids
5.) I'm not sure but I'm pretty sure it's the Doppler effect
6.) Frequency Increases
Answer:
Explanation:
We shall take the help of vector form of displacement . Taking east as i and north as j
4.0m N = 4 j
7.5 m E = 7.5 i
6.8 m S = - 6.8 j
3.7 m E, = 3.7 i
3.6 m S = - 3.6 j
5.3 m W = - 5.3 i
3.7 m N, = 3.7 j
5.6 m W = - 5.6 i
4.4 m S = - 4.4 j
4.9 m W = - 4.9 i
Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i
= -4.6 i -7.1 j
magnitude of displacement = 
= 8.46 m
Direction
Tanθ = 7.1/ 4.6
θ = 57⁰ south of west .
distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9
= 49.5 m
The answer would be 27,000 Joules because (1/2) m v^2 =30*900 which equals 27,000 J
