Answer:
35.7 kg lid we put
Explanation:
given data
temperature = 105 celcius
diameter = 15 cm
Patm = 101 kPa
to find out
How heavy a lid should you put
solution
we know Psaturated from table for temperature is 105 celcius is
Psat = 120.8 kPa
so
area will be here
area = 
..................1
here d is diameter
put the value in equation 1
area = 
area = 0.01767 m²
so net force is
Fnet = ( Psat - Patm ) × area
Fnet = ( 120.8 - 101 ) × 0.01767
Fnet = 0.3498 KN = 350 N
we know
Fnet = mg
mass = 
mass = 
mass = 35.7 kg
so 35.7 kg lid we put
Answer:
Three material considerations are;
1. Identify and appraise the attainment of the goal of the with the design specification
2. Ascertain the required load the product being designed will experience and the suitability of the design specification to that load
3. Review the producibility of the design to ensure that it can be produced with the available technology
Explanation:
1. The appraisal of the design includes the consideration of the factors of the design and the building of reliability and efficiency into the design from the beginning
2. Ascertain if the product will require toughness, elasticity, and if will be subject to sudden or repeated loading conditions
3. Ensure that the design can be readily produced with the accessible manufacturing equipment during the conceptualization stage of the design.
Use protective gear. Use insulated tools, Wear flame resistant clothing, safety glasses, and insulation gloves, Remove watches or other jewelry, Stand on an insulation mat. 03. Never connect the insulation tester to energized conductors or energized equipment and always follow the manufacturer's recommendations. When installing new electrical machinery or equipment, testing insulation resistance is important for two reasons. First, it ensures that the insulation is in adequate condition to begin operation. ... The test is accomplished by applying DC voltage through the de-energized circuit using an insulation tester. Insulation resistance should be approximately one megohm for each 1,000 volts of operating voltage, with a minimum value of one megohm. For example, a motor rated at 2,400 volts should have a minimum insulation resistance of 2.4 megohms.
Answer:
V=L(di/dt) where i is current, V=0.208
Explanation:
using expression iL(t)=5e-2t+3te-2t-2 and L=0.05H(50/1000)
V=0.05*d(5e-2t+3te-2t-2)/dt
since there is no power of e, I'll assume the power to be 1
V=0.05*(-2+3e-2)
at t=0.25
V=0.15e-0.2
V=0.208
