Answer:
radius = 9.1 × 
m
Explanation:
given data
applied load = 5560 N
flexural strength = 105 MPa
separation between the support = 45 mm
solution
we apply here minimum radius formula that is
radius = ![\sqrt[3]{\frac{FL}{\sigma \pi}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7BFL%7D%7B%5Csigma%20%5Cpi%7D%7D)
.................1
here F is applied load and is length
put here value and we get
radius = ![\sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B5560%5Ctimes%2045%5Ctimes%2010%5E%7B-3%7D%7D%7B105%20%5Ctimes%2010%5E6%20%5Cpi%7D%7D)
solve it we get
radius = 9.1 × m
Answer:
Noise or sound is the energy produced by sounding object which can be felt by our hearing organs .
Explanation:
Noise is produced by vibration in a body.
Answer:
Velocity component in x-direction 
.
Explanation:
v=3xy+
y
We know that for incompressible flow
So 
By integrate with respect to x,we will find
+C
So the velocity component in x-direction .
Answer:
(N-1) × (L/2R) = (N-1)/2
Explanation:
let L is length of packet
R is rate
N is number of packets
then
first packet arrived with 0 delay
Second packet arrived at = L/R
Third packet arrived at = 2L/R
Nth packet arrived at = (n-1)L/R
Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R
Now
L / R = (1000) / (10^6 ) s = 1 ms
L/2R = 0.5 ms
average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2
the average queuing delay of a packet = 0 ( put N=1)
