Answer:
Time taken, 
Explanation:
It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.
From the figure,
The sum of forces in y direction is :
Sum of forces in x direction,
.............(1)
Also, 
Equation (1) becomes :
...............(2)
Let t is the time taken for the ball to rotate once around the axis. It is given by :
Put the value of T from equation (2) to the above expression:
On solving above equation :
Hence, this is the required solution.
Given:
u(initial velocity)=0
a=5.54m/s^2
v(final velocity)=2 m/s
v=u +at
Where v is the final velocity.
u is the initial velocity
a is the acceleration.
t is the time
2=0+5.54t
t=2/5.54
t=0.36 sec
It's not so much a "contradiction" as an approximation. Newton's law of gravitation is an inverse square law whose range is large. It keeps people on the ground, and it keeps satellites in orbit and that's some thousands of km. The force on someone on the ground - their weight - is probably a lot larger than the centripetal force keeping a satellite in orbit (though I've not actually done a calculation to totally verify this). The distance a falling body - a coin, say - travels is very small, and over such a small distance gravity is assumed/approximated to be constant.
A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
<span>Answer:
Let m = mass of cannon
Then
10000 = ma
a = 10000/m
v^2 = u^2 + 2as
v^2 = 0 + 2as
84^2 = 2(2.21)(10000/m)
84^2 m = 4.42(10000)
m = 6.264172336
= 6.26 kg
Part 2
Range = u^2sin(2x38)/g
= 84^2sin(76)/9.8
= 698.6129229
= 698.6 m</span>
