Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by


78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa
Answer:
(a) 20 MHz
(b) 1.025 KW
(c) 3.33 ns
(d) 33 pF
Explanation:
(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = <u>20 MHz</u>
(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = <u>1.025 KW</u>
(c) 0.333 x 10^(-8) s = 3.33 x 10^(-9) s = 3.33 nano s = <u>3.33 ns</u>
(d) 33 x10^(-12)F = 33 pico F = <u>33 pF</u>