25,000 Feet = 7620m
PE = mgh where m is mass, g is gravity accel: 9.8 n h is height
= 90 x 9.8 x 7620
= 6720840J
= 6.72MJ
F = ma where m is mass, a is accel = gravity = 9.8
= 90 x 9.8
= 882N
Accel = gravity = 9.8m/s^2
KE = 1/2mv^2 where m is mass n v is vel
if no wind resistance, PE leaving airplane = KE at net
6720840 = 1/2 x 90 x v^2
v^2 = 149352
v = 386.5m/s
Answer:
TEJ as this is a thing you wont get
Lose electrons because they have a negative charge.
Answer:
t = 1.42 s and d = 35.5 m
Explanation:
Given that,
Velocity of a roadrunner is 25 m/s
A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.
We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.
Let d is the distance traveled. So,
d = vt
d = 25 m/s × 1.42 s
d = 35.5 m
<span>The shortening velocity refers to the speed of the contraction from the muscle shortening while lifting a load. Maximal shortening velocity is only attained with a minimal load. With a light load, the shortening velocity is at its Maximal shortening velocity. When the weight is heavy, the speed in which the muscle lifts the weight decreases in speed at a slower velocity.</span>
