<span>orbital velocities to their mean distances from the Sun.</span>
KE=1/2mv^2 - equation for kinetic energy
KE=(1/2)(0.12 kg)((7.8 m/s)^2 - plug it into the formula
KE=(0.06 kg)(60.84 m/s) - multiply 1/2 to the mass and square the speed
KE= 3.7 J - answer
Hope this helps
1 astronomical unit = 149597870700m
Enrico should divide distance in meters with this number.
A projectile motion is characterized by motion moving in a direction of an arc. It is acted upon by two component vectors: the horizontal and vertical. These two vectors are independent of each other when it comes to time of flight. The horizontal direction travels at constant speed, while the vertical direction travels at constant acceleration due to gravity, The time for an object to reach the ground would be equal, whether dropped from the sampe point or thrown in a projectile motion. Of course, this is assuming ideality wherein there is no air resistance.
So, the hang up time, or the time the object stayed on air is calculated using this equation:
a = Δv/t
Δv is the change in velocity which is the initial velocity when it was dropped to when it reaches zero velocity when it hits the ground.
9.81 m/s² = |(0 - 7.3)|/t
t = 0.744 seconds
Given :
Walk in forward direction is 30 m .
Walk in backward direction is 25 m .
To Find :
The distance and displacement .
Solution :
We know , distance is total distance covered and displacement is distance between final and initial position .
So , distance travelled is :
D = 30 + 25 m = 55 m .
Now , we first move 30 m in forward direction and then 25 m in backward direction .
So , displacement is :
D = 30 - 25 m = 5 m .
Therefore , distance and displacement covered is 55 m and 5 m respectively .
Hence , this is the required solution .
