Highest elevation is 349
Contour interval is 50
Answer:
the tension in each side of the cable is 3677.57 N
Explanation:
given data
traffic light = 20 kg
cable between two poles = 30 m
sag in the cable = 0.40 m
solution
by the free body diagram
tan θ = 
.............1
θ = 1.527 °
and
tension = mg
The net force is along x - axis is express as
T2 cosθ = T1 cosθ .................2
so T2 - T1 ..............3
and
when we take it along y axis that is express as
( T1 + T2) sinθ = mg ...................4
so by equation 3 we put here
2 × T1 sin(1.527) = 20 × 9.8
T1 = 3677.57 N
Answer:
Explanation:
Descriptive statistics statistics are measures that summarize the characteristics of a sample.
Hope it helps
<span> we suppose that x be distance of man from spot light is
so 12-x is distance from man to wall
we will draw first triangle ABC, where
a= spotlight (on ground)
b= man (feet)
c= man (head)
we know that AB = x and BC = 2
by extend line AB to the wall at point D and extend line AC to the wall at point E
AD = 12 (distance from spotlight to wall)
DE = s (length of shadow)
Now triangles ABC and ADE are similar
Therefore DE/AD = BC/AB
s / 12 = 2 / x
s = 24 / x
Differentiate both sides with respect to t
ds/dt = -24/x² dx/dt
Man is walking toward building at speed of 16 ms
dx/dt = 16
Man is 4 m from the building/wall
12 - x = 4
x = 8
Find ds/dt when x = 8 and dx/dt = 16
ds/dt = -24/x² dx/dt
ds/dt = -24/(64) * 16
ds/dt = -6
So length of shadow is decreasing at rate of 6 m/s
hope it helps
</span>
<span>protection from injustices</span>
