Answer: wear, safety goggles, lab coat, gloves, hair tied back, and all jewelry off. Make sure to have someone in the room like a teacher in case someone gets hurt, I hope this helps:)
Explanation:
Answer:
- Driving surface area as subject to differential pressures
- How to achieve continuous rotation
- Mechanical connections
Explanation:
The general stages of thinking when designing a system are;
- First understanding the problem that requires you to design a solution
- Defining the problem that needs to be solved. In this case is to design an efficient hydraulic engine for a motorboat
- Research to find any available solutions for similar problems
- Idea of the solution
- A prototype to try solve the challenge
- Selecting and adjusting the prototype as you implement the solution.
For a hydraulic engine design, the following designing parts are vital;
- Driving surface area as subject to differential pressures
- How to achieve continuous rotation
- Mechanical connection
Pressures are important due to motor displacements. Fuel burns in the cylinder to cause power through fuel gases burning and expansion which is the four-stroke cycle. Continuous rotation will come in the design considerations through reciprocating motions of pistons into rotary motions. Mechanical connections are vital especially in crank shaft turns, gear box that turns horizontal motions to spinning motions and finally the propeller that drives the boat on the water.
Answer:
Explanation:
PE = Energy of food = 500 cal = 
m = Mass of object = 50 kg
g = Acceleration due to gravity = 
Potential energy of food is given by
Nancy could raise the weight to a maximum height of .
Mass of 
used per year = 
Energy of = 
Power
The power requirement is .
Answer:
a. 81 kj/kg
b. 420.625K
c. 101.24kj/kg
Explanation:
![\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7Bt1%7D%20%3D%5B%5Cfrac%7Bp2%7D%7Bp1%7D%20%5D%5E%7B%5Cfrac%7By-1%7D%7By%7D%20%7D)
t1 = 360
p1 = 0.4mpa
p2 = 1.20
y = 1.13
substitute these values into the equation
![\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.20%7D%7B0.4%7D%20%5D%5E%7B%5Cfrac%7B1.13-1%7D%7B1.13%7D%20%7D)
![\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.2%7D%7B0.4%7D%20%5D%5E%7B0.1150%7D%5C%5C%5Cfrac%7Bt2%7D%7B360%7D%20%3D1.1347)
when we cross multiply
t2 = 360 * 1.1347
= 408.5
a. the work required in the firs compressor
w=c(t2-t1)
c=1.67x10³
t1 = 360
t2 = 408.5
w = 1670(408.5-360)
= 1670*48.5
= 80995 J
= 81KJ/kg
b. 
n = 80%
t2 = 408.5
t1 = 360
0.80 = 408.5-360 ÷ t'2-360
cross multiply to get the value of t'2
0.80(t'2-360) = 48.5
0.80t'2 - 288 = 48.5
0.8t'2 = 48.5+288
0.8t'2 = 336.5
t'2 = 336.5/0.8
= 420.625
this is the temperature at the exit of the first compressor
c. cooling requirement
w = c(t2-t1)
= 1.67x10³(420.625-360)
= 1670*60.625
= 101243.75
= 101.24kj/kg
Leading/trailing shoe type drum brake This is called the servo effect (self-boosting effect) which realizes the powerful braking forces of drum brakes. ... This is because drum brakes generate the same braking force in either direction. Generally, this type is used for the rear brakes of passenger cars.
