Answer:
0.78m (rounded to nearest hundredth of a meter)
explanation:
time taken for going up=time taken for drop down after reaching the highest point. at the highest point, the velocity becomes 0.
now all thats left is dropping an object from a height (h) and seeing how long it takes to reach the ground. then find out the flight’s total time divided by 2 (0.8/2=0.4)
lets say the velocity is v and the height she jumped to is h. we can make a kinematic expression:
s=vt+½gt²
once we put it all together you should get this:
h=0×0.4+½(9.81) 0.4²
.
∴
Time taken for downward drop
=
0.8
2
=
0.4
s
Suppose that she jumped with initial velocity
=
u
Also suppose that she jumped to a height
h
Using following kinematic expression
s
=
u
t
+
1
2
g
t
2
and inserting various quantities we get
h
=
0
×
0.4
+
1
2
(
9.81
)
0.4
2
h
=
0.78
m
rounded to nearest hundredth of a meter.
The answer for this question should be TRUE
-- Before Adrian left the airplane, his gravitational potential energy was
(mass) x (gravity) x (height) = (80kg) x (9.81m/s²) x (1,000m) = 784,800 joules
-- When he reached the ground, his kinetic energy was
(1/2) x (mass) x (speed)² = (40kg) x (5m/s)² = 1,000 joules
-- Between the airplane and the ground, the Adrian lost
(784,800 joules) - (1,000 joules) = 783,800 joules
Where did all that energy go ?
Energy never just disappears. If it's missing, it had to go somewhere.
The Adrian used 783,800 joules of energy to push air our of his way
so that he could continue his parachute jump, and reach the ground
in time to be home for dinner.
The answer is C radio waves are very high energy mechanical waves.
Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.
In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.
Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.
We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.
