The load is 17156 N.
<u>Explanation:</u>
First compute the flexural strength from:
σ = FL / π
= 3000 
(40 10^-3) / π (5 10^-3)^3
σ = 305 10^6 N / m^2.
We can now determine the load using:
F = 2σd^3 / 3L
= 2(305 10^6) (15 10^-3)^3 / 3(40 10^-3)
F = 17156 N.
Answer:
6.99 x 10⁻³ m³ / s
Explanation:
Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.
h = difference in the height of water column at two ends of delivery pipe
6 - 1 = 5 m
Velocity of flow of water
v = √2gh
= √ (2 x 9.8 x 5)
= 9.9 m /s
Volume of water flowing per unit time
velocity x cross sectional area
= 9.9 x 3.14 x .015²
= 6.99 x 10⁻³ m³ / s
Radio waves are radiated by charged particles when they are accelerated. They are produced artificially by time-varying electric currents, consisting of electrons flowing back and forth in a specially-shaped metal conductor called an antenna. ... Radio waves are received by another antenna attached to a radio receiver.
Answer:
attached below is the detailed solution and answers
Explanation:
Attached below is the detailed solution
C(iii) : versus the parameter C
The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C
Answer:
D
Explanation: She hopes to be able to make this, however she hasn't yet...therefore she is thinking of a concept and it's development
