Answer:
![l = 250.3\,mm](https://tex.z-dn.net/?f=l%20%3D%20250.3%5C%2Cmm)
Explanation:
The size needed to use the kerf on the laser properly is:
![l = 250\,mm + 0.3\,mm](https://tex.z-dn.net/?f=l%20%3D%20250%5C%2Cmm%20%2B%200.3%5C%2Cmm)
![l = 250.3\,mm](https://tex.z-dn.net/?f=l%20%3D%20250.3%5C%2Cmm)
Answer: r = 0.8081; s = -0.07071
Explanation:
A = (150i + 270j) mm
B = (300i - 450j) mm
C = (-100i - 250j) mm
R = rA + sB + C = 0i + 0j
R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j
R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j
Equating the i and j components;
150r + 300s - 100 = 0
270r - 450s - 250 = 0
150r + 300s = 100
270r - 450s = 250
solving simultaneously,
r = 0.8081 and s = -0.07071
QED!
Answer:
150 N
Explanation:
Moment is the product of force and distance:
27 N·m = F·(0.180 m)
F = 27 N·m/(0.180 m) = 150 N
The force required to produce the desired moment is 150 N.
Answer:
471 days
Explanation:
Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons
As,
1 gallon = 0.133 cubic feet (cf)
Therefore,
Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133
= 4.28 x 10˄8
Applying Mass balance i.e
Accumulation = Mass In - Mass out (Eq. 01)
Here
Mass In = 0.5 cfs
Mass out = 11 cfs
Putting values in (Eq. 01)
Accumulation = 0.5 - 11
= - 10.5 cfs
Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.
Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600
= 37,800
Converting depletion of reservoir in cubic feet per day = 37, 800 x 24
= 907,200
i.e. 907,200 cubic feet volume is being depleted in days = 1 day
1 cubic feet volume is being depleted in days = 1/907,200 day
4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200
= 471 Days.
Hence in case of continuous drought reservoir will last for 471 days before dry-up.
Answer: heat flux into the fun is 21.714 mW/m^2
Explanation:
Heat flux Q = q/A
q = heat transfer rate W
A = area m^2
q = area * conductivity * temperature gradient
Temperature gradient = difference in temperature of the metal faces divided by the thickness.
Therefore Q = k * ( temp. gradient)
Q = 200 * ((400-20)/3.5*10^-2)
Q = 21714285.71 = 21.714 mW/m^2
Answer 2: convective heat transfer flux between fin and air
is 3800W/m^2
Explanation :
q = hA*(Ts-Ta)
h = convective heat transfer coefficient
Ts = temperature of fin
Ta = temperature of air
Q = q/A
Q = h(Ts-Ta)
Q = 10(400 - 20)
Q = 3800 W/m^2