The speed of the earth's surface located at 2/3 of the length of the arc between the pole which measure from the equator is 232.5 m/s.
Solution:
So the givens are, earth's radius = 6.37X10^6m, and the angular distance from the pole is 90 degrees. So 60 degrees is the 2/3.
r = 6.37x10^6 * cos(60) = 3.185x10^6m
since v = wr
v = 7.3x10^-5 * 3.185x10^6
v - 232.5 m/s
Answer:
1.944m_s²
Explanation:
First convert speed from km/h to m/s the use the formula a=v-u
t
Explanation:
s = ut + 1/2 a t^2
375 = 0 * 5 + 1/2 * a * (5)^2
375 = 1/2 * a * 25
a = 375*2/25
a = 15* 2
a = 30m/sec^2
v = u + at
v = 0 + 30 * 5
v = 150 m/sec
hope it helps you
C that is the condensation point
The maximum allowable torque must correspond to the allowable shear stress for maximization. To solve this, we use the torsion formula:
Max. Allowable Shear Stress = Maximum Torque ÷ Cross-Sectional Area
8 x 10^6 Pa = Maximum Torque ÷ pi*(d/2)²
Maximum Torque = 8 x 10^6 Pa * pi*(0.06/2)² m²
Maximum Torque = 22,619.47 J or
Maximum Torque = 22.62 kJ
As for the second question, I have no reference figure so I am unable to answer it. I hope I was still able to help you, though.
