Answer:
V = 0.5 m/s
Explanation:
given data:
width of channel = 4 m
depth of channel = 2 m
mass flow rate = 4000 kg/s = 4 m3/s
we know that mass flow rate is given as
![\dot{m}=\rho AV](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%3D%5Crho%20AV)
Putting all the value to get the velocity of the flow
![\frac{\dot{m}}{\rho A} = V](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cdot%7Bm%7D%7D%7B%5Crho%20A%7D%20%3D%20V)
![V = \frac{4000}{1000*4*2}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4000%7D%7B1000%2A4%2A2%7D)
V = 0.5 m/s
Answer:
Headlights are required to be used 1/2 hour after sunset to 1/2 hour before sunrise, when windshield wipers are being used, when visibility is less than 1000 feet, or when there is insufficient light or adverse weather.
Explanation:
hope this helps
Answer:
D. a triangle and a T-Square
Explanation:
A T-Square is the best drawing tool to create squares. You would need a squares to create cubes.
Answer:
Both Technician A and B are correct
Explanation:
Due to a contamination of the clutch disc friction surface, the clutch chatters
The contamination that causes clutch chattering includes
1) Worn or broken motor mounts
2) Bell housing bolts becoming loose
3) Clutch link damage
4) Warped flywheel, due to overheating, the flywheel can become warped such that the non uniform surface interfaces the clutch resulting in clutch chattering
5) Engine or transmission oil contaminating the disc.
Answer:
a) 2,945 mC
b) P(t) = -720*e^(-4t) uW
c) -180 uJ
Explanation:
Given:
i (t) = 6*e^(-2*t)
v (t) = 10*di / dt
Find:
( a) Find the charge delivered to the device between t=0 and t=2 s.
( b) Calculate the power absorbed.
( c) Determine the energy absorbed in 3 s.
Solution:
- The amount of charge Q delivered can be determined by:
dQ = i(t) . dt
![Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt](https://tex.z-dn.net/?f=Q%20%3D%20%5Cint%5Climits%5E2_0%20%7Bi%28t%29%7D%20%5C%2C%20dt%20%3D%20%5Cint%5Climits%5E2_0%20%7B6%2Ae%5E%28-2t%29%7D%20%5C%2C%20dt%20%3D%206%2A%5Cint%5Climits%5E2_0%20%7Be%5E%28-2t%29%7D%20%5C%2C%20dt)
- Integrate and evaluate the on the interval:
![= 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C](https://tex.z-dn.net/?f=%3D%206%20%2A%20%28-0.5%29%2Ae%5E-2t%20%3D%20-%203%2A%28%201%20%2F%20e%5E4%20-%201%29%20%3D%202.945%20C)
- The power can be calculated by using v(t) and i(t) as follows:
v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt
v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV
P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)
P(t) = -720*e^(-4t) uW
- The amount of energy W absorbed can be evaluated using P(t) as follows:
![W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%5Climits%5E3_0%20%7BP%28t%29%7D%20%5C%2C%20dt%20%3D%20%5Cint%5Climits%5E2_0%20%7B-720%2Ae%5E%28-4t%29%7D%20%5C%2C%20dt%20%3D%20-720%2A%5Cint%5Climits%5E2_0%20%7Be%5E%28-4t%29%7D%20%5C%2C%20dt)
- Integrate and evaluate the on the interval:
![W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ](https://tex.z-dn.net/?f=W%20%3D%20-180%2Ae%5E-4t%20%3D%20-%20180%2A%28%201%20%2F%20e%5E12%20-%201%29%20%3D%20-180uJ)