Answer:
2420 J
Explanation:
From the question given above, the following data were obtained:
Force (F) = 22.9 N
Angle (θ) = 35°
Distance (d) = 129 m
Workdone (Wd) =?
The work done can be obtained by using the following formula:
Wd = Fd × Cos θ
Wd = 22.9 × 129 × Cos 35
Wd = 22.9 × 129 × 0.8192
Wd ≈ 2420 J
Thus, the workdone is 2420 J.
Answer:
a)The approximate radius of the nucleus of this atom is 4.656 fermi.
b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Explanation:

= Constant for all nuclei
r = Radius of the nucleus
A = Number of nucleons
a) Given atomic number of an element = 25
Atomic mass or nucleon number = 52


The approximate radius of the nucleus of this atom is 4.656 fermi.
b) 
k=
= Coulombs constant
= charges kept at distance 'a' from each other
F = electrostatic force between charges


Force of repulsion between two protons on opposite sides of the diameter



The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is
H = 50 - (1/2 G Q²) ,
and the height of the stone that was tossed straight up
from the ground is
H = 20Q - (1/2 G Q²) .
The stones meet when them's heights are equal,
so that's the time when
<span>50 - (1/2 G Q²) = 20Q - (1/2 G Q²) .
This is looking like it's going to be easy.
Add </span><span>(1/2 G Q²) to each side.
Then it says
50 = 20Q
Divide each side by 20: 2.5 = Q .
And there we are. The stones pass each other
2.5 seconds
after they are simultaneously launched.
</span>
Answer:
x = 2.044 m
Explanation:
given data
initial vertical component of velocity = Vy = 2sin18
initial horizontal component of velocity = Vx = 2cos18
distance from the ground yo = 5m
ground distance y = 0
from equation of motion


solving for t
t = 1.075 sec
for horizontal motion

x = 2cos18*1.075
x = 2.044 m
Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that ball will reach at maximum height at
t = 3 s
now we will have

now we have


Now maximum height above ground is given as



Part b)
Height of the fence is given as



Part c)
As we know that its horizontal distance moved by the ball in 5.5 s is given as



now total time of flight is given as

so range is given as



so the distance from the fence is given as

