Answer:
0.0072 m³/s
Explanation:
Using Bernoulli's law
P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂ since the pipe is horizontal
1/2ρv₂² - 1/2ρv₁² = P₁ - P₂
flow rate is constant
A₁v₁ = A₂v₂
A₁ = πr₁² = π (0.06/2)² = 0.0028278 m²
A₂ = πr₂² = π (0.0225)² = 0.00159 m²
v₁ = (A₂ / A₁)v₂
v₁ = (0.00159 m²/ 0.0028278 m²) v₂ = 0.562 v₂
substitute v₁ into the Bernoulli's equation
1/2ρv₂² - 1/2ρv₁² = P₁ - P₂
500 ( 1 - 0.3161 ) v₂² = (31.0 - 24 ) × 10³ Pa
341.924 v₂² = 7000
v₂² = 20.472
v₂ = √ 20.472 = 4.525 m/s
volume follow rate = 0.00159 m² × 4.525 m/s = 0.0072 m³/s
Answer:
![x=2.4365\ m](https://tex.z-dn.net/?f=x%3D2.4365%5C%20m)
and
![x=-1.4365\ m](https://tex.z-dn.net/?f=x%3D-1.4365%5C%20m)
Explanation:
Given:
- first charge,
![q_1=5\times 10^{-3}\ C](https://tex.z-dn.net/?f=q_1%3D5%5Ctimes%2010%5E%7B-3%7D%5C%20C)
- second charge,
![q_2=3\times 10^{-3}\ C](https://tex.z-dn.net/?f=q_2%3D3%5Ctimes%2010%5E%7B-3%7D%5C%20C)
- position of first charge,
![x_1=-2\ m](https://tex.z-dn.net/?f=x_1%3D-2%5C%20m)
- position of second charge,
![x_2=-1\ m](https://tex.z-dn.net/?f=x_2%3D-1%5C%20m)
Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.
<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>
![E_1=E_2](https://tex.z-dn.net/?f=E_1%3DE_2)
- since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.
![\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5Cpi.%5Cepsilon_0%7D%20%5Cfrac%7Bq_1%7D%7B%28r%2B1%29%5E2%7D%20%3D%5Cfrac%7B1%7D%7B4%5Cpi.%5Cepsilon_0%7D%20%5Cfrac%7Bq_2%7D%7B%28r%29%5E2%7D)
![\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}](https://tex.z-dn.net/?f=%5Cfrac%7B5%5Ctimes%2010%5E%7B-3%7D%7D%7B%28r%2B1%29%5E2%7D%20%3D%20%5Cfrac%7B3%5Ctimes%2010%5E%7B-3%7D%7D%7B%28r%29%5E2%7D)
![3(r^2+1+2r)=5r^2](https://tex.z-dn.net/?f=3%28r%5E2%2B1%2B2r%29%3D5r%5E2)
![2r^2-6r-3=0](https://tex.z-dn.net/?f=2r%5E2-6r-3%3D0)
![r=3.4365 \&\ r=-0.4365](https://tex.z-dn.net/?f=r%3D3.4365%20%5C%26%5C%20r%3D-0.4365)
Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.
![x=-1+3.4365=2.4365\ m](https://tex.z-dn.net/?f=x%3D-1%2B3.4365%3D2.4365%5C%20m)
and
![x=-1-0.4365=-1.4365\ m](https://tex.z-dn.net/?f=x%3D-1-0.4365%3D-1.4365%5C%20m)
Answer:
O D.
Explanation:
Physics has an aspect that deals with the study of energy