Answer:
a. 4
b. 1 m
Explanation:
According to the question, the data is as follows
The Density of water at 20 degrees celcius is 1000 kg/m^3
Viscosity is 0.001kg/m/.s
Velocity V = 25 cm/s
V = 0.25 m/s
Now
a. The creeping motion is
As we know that
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
1 = (1,000 × 0.25 × d) ÷ 0.0001
d = (1 × 0.001) ÷ (1,000 × 0.25)
= 4E - 06^m
= 4
b. Now the sphere diameter is
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
250,000 = (1,000 × 0.25 × d) ÷ 0.0001
d = (250,000 × 0.001) ÷ (1,000 × 0.25)
= 1 m
Answer:
Correct option: B. 90%
Explanation:
The confidence interval is given by:
If is 190, we can find the value of
:
Now we need to find the value of :
So the value of z is 1.71.
Looking at the z-table, the z value that gives a z-score of 1.71 is 0.0436
This value will occur in both sides of the normal curve, so the confidence level is:
The nearest CI in the options is 90%, so the correct option is B.
Answer:
Qcd=0.01507rad
QT= 0.10509rad
Explanation:
The full details of the procedure and answer is attached.
Answer:
final temperature is 424.8 K
so correct option is e 424.8 K
Explanation:
given data
pressure p1 = 1 bar
pressure p2 = 5 bar
index k = 1.3
temperature t1 = 20°C = 293 k
to find out
final temperature t2
solution
we have given compression is reversible and has an index k
so we can say temperature is
...........1
put here all these value and we get t2
t2 = 424.8
final temperature is 424.8 K
so correct option is e