Answer:
3.49 seconds
3.75 seconds
-43200 ft/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration

Time the parachutist falls without friction is 3.19 seconds

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds


Now the initial velocity of the last half height will be the final velocity of the first half height.

Since the height are equal


Time taken to fall the first half is 2.65 seconds
Total time taken to fall is 2.65+1.1 = 3.75 seconds.
When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

Magnitude of acceleration is -43200 ft/s²
No the substance will remain the same substance as before.
It is D because our eye lenses reflect the white light we see and it also reflects the light to a point to where we can see colors and objects clearly... Hope this helps out ^-^''
Lol i think someone would be fired from their job if they threw food
Answer:
a) T=1.35s
b) amplitude = 0.0923m
Explanation:
m=300 gr
k=6.5 N/m
first we need to get the angular frequency of the motion
so we have that
ω = √(k/m)
in this case motion is a simple harmonic so the period is defined by:
T= 2π / ω
T= 2π / √(k/m)
replacing the variables...
T= 2π / √(6.5/0.3)
T=1.35s (period of the block's motion)
and...
α max = | ω²r max |
2 = (2π/1.35)² * r max
r max= 0.0923m