Answer:
Check attachment for better understanding
Explanation:
Given that
a= 0.75m
b=1.31m
c= 2.2m
Weight of pole is 26.90
Then, Fg = Weight = 26.90
Using Equilibrium of forces
ΣFy = 0
U — D — Fg = 0
U — D = Fg
U — D = 26.9
To calculate U,
We will take moment about point A.
ΣMa = 0
Let the clockwise moment be positive and anti-clockwise be negative
Fg(a+b) — U(a) = 0
26.9(1.31+0.75) —0.75U = 0
26.9(2.06) = 0.75U
0.75U = 55.414
U = 55.414/0.75
U = 73.89 N
To calculate D,
U — D = 26.9
73.89—D =26.9
73.89—26.9 = D
D = 46.99N
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Answer:
If energy is conserved, then the sum of the potential energy and the kinetic energy is a constant.
Assuming the proton starts from rest, so it's kineitc energy is zero, but it has a potential energy, PE equal to:
PE = qV
where q =1.6 x 10^-19 C
and V = 1.00 V
Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:
PE* = 0,
KE* = 1/(2mv^2)
Now since
PE + KE = Total energy =PE* + KE*
Therefore,
qV + 0 = 0 + 1/2mv^2
Or
KE = qV = 1.6 10^-19 J
