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
Answer:
The velocity is 
Explanation:
From the question we are told that
The first distance is 
The first speed is 
The second distance is 
The second speed is 
Generally the time taken for first distance is
The time taken for second distance is
The total time is mathematically represented as
=> 
=> 
Generally the constant velocity that would let her finish at the same time is mathematically represented as
=> 
=>
Apply Newton's second law to the person's motion:
F = ma
F = net force, m = mass, a = acceleration
Given values:
m = 50.8kg, a = 3.50m/s²
Plug in and solve for F:
F = 50.8(3.50)
F = 178N
Answer:
Period of the signal.
Explanation:
So, this question is all about a concept in physics or astronomy which is called or known as Radiation Astronomy and Galactic Nuclei that are active. This concept talks most about Quasars; a powerful radiating object which derives its power from black holes.
When You take a look at Quasars, we get the to know that the more you think you can see, the more they move away from us.
Thus, when "You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength regularly over a certain period. The maximum possible size for the source of this radiation can now be calculated from the "PERIOD OF THE SIGNAL.
NB: not the amplitude but the period.
Answer:
too much exposure to the sun's rays
