Have a universal record base. Everyone is able to understand the data compiled since the same measurement systems are being used around the world. This is just to simplify all of the information.
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.
is the acceleration of the box.
<u>Explanation:</u>
Given data:
Mass of the box = 3.74 kg
Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.
First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below. Net horizontal force acting on the box (F) is given by
F = 2.699676 – 1.192275 = 1.507 N
Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as
Answer:
0.0002
Explanation:
First you have to solve exponents first, so you solve 10 to the power of negative 4 which is 0.0001 and now you have 2.0x0.0001 and that equals 0.0002
