By equation of motion we have v = u + at
Where u = Initial velocity, v = final velocity, t = time taken and a = acceleration
Here v = 141 m/s, u = 17.7 m/s and t = 6 s
On substitution we will get
141 = 17.7+ 6a
So, a = (141-17.7)/6 = 20. 55 m/
Aceeleration = 20. 55 m/
along north direction.
Answer:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.
Answer:
η = 58.8%
Explanation:
Work is defined as the force applied by the distance traveled by the body.

where:
W = work [J] (units of joules)
F = force = 294 [N]
d = distance = 5 [m]
![W = 294*5\\W = 1470 [J]\\](https://tex.z-dn.net/?f=W%20%3D%20294%2A5%5C%5CW%20%3D%201470%20%5BJ%5D%5C%5C)
Efficiency is defined as the energy required to perform an activity in relation to the energy actually added to perform some activity. This can be better understood by means of the following equation.

Answer:
160N
Explanation: When 80kg mass is one group . It's reaction force acting on a ground.
Weight of the object = 80*10
= 800 N
Here we are given cofficient of static friction its 0.2. It should be smaller than 1
Friction force = Reaction * Friction Cofficient
Reaction = 800N ( Considering Vertical Equilibrium )
F = 800* 0.2
F = 160N