Answer:
H_w = 2.129 m
Explanation:
given,
Width of the weir, B = 1.2 m
Depth of the upstream weir, y = 2.5 m
Discharge, Q = 0.5 m³/s
Weir coefficient, C_w = 1.84 m
Now, calculating the water head over the weir
![Q = C_w BH^{3/2}](https://tex.z-dn.net/?f=Q%20%3D%20C_w%20BH%5E%7B3%2F2%7D)
![H = (\dfrac{Q}{C_wB})^{2/3}](https://tex.z-dn.net/?f=H%20%3D%20%28%5Cdfrac%7BQ%7D%7BC_wB%7D%29%5E%7B2%2F3%7D)
![H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}](https://tex.z-dn.net/?f=H%20%3D%20%28%5Cdfrac%7B0.5%7D%7B1.84%5Ctimes%201.2%7D%29%5E%7B2%2F3%7D)
![H = 0.371\ m](https://tex.z-dn.net/?f=H%20%3D%200.371%5C%20m)
now, level of weir on the channel
H_w = y - H
H_w = 2.5 - 0.371
H_w = 2.129 m
Height at which weir should place is equal to 2.129 m.
Answer= 8m/s
Because total Momentum before= total momentum after
Momentum before (p=mu)
p=(4)(12)= 48
p=2(0)=0
So total momentum before=48
Momentum after (p=mu)
Masses combined —2+4=6kg
p=6u
Mb=Ma
48=6u
u=8m/s
Answer:
change in entropy is 1.44 kJ/ K
Explanation:
from steam tables
At 150 kPa
specific volume
Vf = 0.001053 m^3/kg
vg = 1.1594 m^3/kg
specific entropy values are
Sf = 1.4337 kJ/kg K
Sfg = 5.789 kJ/kg
initial specific volume is calculated as
![v_1 = vf + x(vg - vf)](https://tex.z-dn.net/?f=v_1%20%3D%20vf%20%2B%20x%28vg%20-%20vf%29)
![= 0.001053 + 0.25(1.1594 - 0.001053)](https://tex.z-dn.net/?f=%3D%200.001053%20%2B%200.25%281.1594%20-%200.001053%29%20)
![v_1 = 0.20964 m^3/kg](https://tex.z-dn.net/?f=v_1%20%3D%200.20964%20%20m%5E3%2Fkg)
![s_1 = Sf + x(Sfg)](https://tex.z-dn.net/?f=s_1%20%3D%20Sf%20%2B%20x%28Sfg%29)
![= 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K](https://tex.z-dn.net/?f=%3D%201.4337%20%2B%200.25%20%5Ctimes%205.7894%20%3D%202.88%20kJ%2Fkg%20K)
FROM STEAM Table
at 200 kPa
specific volume
Vf = 0.001061 m^3/kg
vg = 0.88578 m^3/kg
specific entropy values are
Sf = 1.5302 kJ/kg K
Sfg = 5.5698 kJ/kg
constant volume so![v_1 - v_2 = 0.29064 m^3/kg](https://tex.z-dn.net/?f=%20v_1%20-%20%20v_2%20%20%3D%200.29064%20m%5E3%2Fkg)
![v_2 = v_1 = vf + x(vg - vf)](https://tex.z-dn.net/?f=v_2%20%3D%20v_1%20%3D%20vf%20%2B%20x%28vg%20-%20vf%29)
![=0.29064 = x_2(0.88578 - 0.001061)](https://tex.z-dn.net/?f=%3D0.29064%20%3D%20x_2%280.88578%20-%200.001061%29)
![x_2 = 0.327](https://tex.z-dn.net/?f=x_2%20%3D%200.327)
![s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K](https://tex.z-dn.net/?f=s_2%20%3D%201.5302%20%2B%200.32%20%5Ctimes%205.5968%20%3D%203.36035%20kJ%2Fkg%20K)
Change in entropy ![\Delta s = m(s_2 - s_1)](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%20m%28s_2%20-%20s_1%29)
=3( 3.36035 - 2.88) = 1.44 kJ/kg
The correct option is (D) Gamma (
![_0^0\gamma](https://tex.z-dn.net/?f=_0%5E0%5Cgamma)
)
Explanation:Now this question is a tricky one because all of these options are somehow involved in radioactive decay; however, in this case the SYMBOL is required NOT the elements. There are three symbols involved in radioactive decay, which are:
1. α for alpha decay
2. β for beta decay
3. γ for gamma decay
In the options only one symbol is present which is
gamma. Hence option (D) Gamma ( ![_0^0\gamma](https://tex.z-dn.net/?f=_0%5E0%5Cgamma)
) is the correct answer.
-i
Answer:
The alligator will take t = 10 s to reach the final speed of 35 m/s
Explanation:
As we know that the initial speed of the alligator is 5 m/s
then it accelerate by given acceleration to reach the final speed of 35 m/s
so we will have
![v_i = 5 m/s](https://tex.z-dn.net/?f=v_i%20%3D%205%20m%2Fs)
![v_f = 35 m/s](https://tex.z-dn.net/?f=v_f%20%3D%2035%20m%2Fs)
![a = 3m/s^2](https://tex.z-dn.net/?f=a%20%3D%203m%2Fs%5E2)
now we have
![v_f = v_i + at](https://tex.z-dn.net/?f=v_f%20%3D%20v_i%20%2B%20at)
![35 = 5 + 3 t](https://tex.z-dn.net/?f=35%20%3D%205%20%2B%203%20t)
![t = 10 s](https://tex.z-dn.net/?f=t%20%3D%2010%20s)