We can use Newton II here (where F=m*a), that F is the net (or resultant) force on the object, m is the mass of the object and a is the acceleration the object experiences.
This means, in this case there would be no friction and absolutely no other force which gives a component in the plane of motion, only then can you assume that F=804N.
Now using F= m*a
804 = 51.7*a
Therefore a = 804/51.7 = 15.55 m/s²
We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as
Distance = speed x time
From the information given,
speed = 320 m/s
time = 4.5 s
By substituting these values into the formula, we have
Distance = 320 m/s x 4.5s
s cancels out. We are left with m. Thus,
Distance = 1440m
Answer:
An object has potential energy (stored energy) when it is not in motion. Once a force has been applied or it begins to move the potential energy changes to kinetic energy (energy of motion).
EXAMPLE: A rock sitting on the edge of a cliff. If the rock falls, the potential energy will be converted to kinetic energy, as the rock will be moving. A stretched elastic string in a longbow.
Explanation:
initial height, yo = 2 m
initial velocity, u = 20 m/s
angle of projection,θ = 5 degree
distance of net = 7 m
height of net = 1 m
Let it covers a vertical distance y in time t .
Use Second equation of motion for vertical motion
As it hits the ground in time t, so put y = 0
Taking positive sign, t = 0.84 s
The ball travels a horizontal distance x in time t
X = 20 Cos5 x t
X = 16.76 m
As this distance is more than the distance of net, so it clears the net.
Let t' be the time taken to travel a horizontal distance equal to the distance of net
7 = 20 cos5 x t'
t' = 0.35 s
Let the vertical distance traveled by the ball in time t' is y'.
So,
y' = 2.008 m
So, it clears the net which is 1 m high.
It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m
your welcome, and have a great day.
