Answer:
Explanation:
If a baseball is hit into the air with a velocity of 27 m/s, we want to determine the maximum height of the ball. Using the projecile formula;
Max height H = u²/2g
u is the initial velocity of the body = 27m/s
g is the acceleration due to gravity = 9.81m/s²
H = 27²/2(9.81)
H = 729/19.62
H = 37.16m
Hence the ball went 37.16m high
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:
It is formed by a horizontal number line, called the x-axis, and a vertical number line, called the y-axis.
Explanation:
Answer:
The magnitude and direction are
7.638×10-4m
80.01°
Explanation:
We know that the coefficient of linear expansion for copper = 16.6×10^-6 m/m-C
ΔT = 40.2 - 18.0 = 28.5 C°
The expansion of horizontal pipe length can be calculated as
= (0.28)(16.6×10^-6)(28.5) = 13247×10^-8
= 0.0001325 m
The expansion of vertical pipe length = (1.28)(16.6×10^-6)(28.5) = 60557×10^-8 = 0.000752229 m
horizontal displacement = 0.1325 mm
= 1.356×10^-4m
vertical displacement = 0.75223mm
=7.5223×10-4m
size of total displacement can be calculated as
√(x²+y²)
Where x and y are vertical and horizontal displacement respectively
= √(0.1325)²+(0.75223)² =
= 0.7638 mm
= 7.638×10-4m
Angle below horizontal = arctan Θ
= 0.75223/0.1325
=5.6772
= arctan (5.6772)
= 80.01°
Therefore, the the magnitude and direction of the displacement of the pipe elbow when the water flow is turned at (7.638×10-4m) 0.7638 mm and 80.01°