Answer:
R min = 28.173 ohm
R max = 1.55 × ohm
Explanation:
given data
capacitor = 0.227 μF
charged to 5.03 V
potential difference across the plates = 0.833 V
handled effectively = 11.5 μs to 6.33 ms
solution
we know that resistance range of the resistor is express as
V(t) = ...........1
so R will be
R = ....................2
put here value
so for t min 11.5 μs
R =
R min = 28.173 ohm
and
for t max 6.33 ms
R max =
R max = 1.55 × ohm
Question in order:
A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?
Mirror Radius (mm) Bending Failure Stress (MPa)
0.603 225
0.203 368
0.162 442
Answer:
191 MPa
Explanation:
Failure stress of bending is Inversely proportional to the mirror radius
At mirror radius 1 = 0.603 mm Bending stress 1 = 225 Mpa..............(1)
At mirror radius 2 = 0.203 mm Bending stress 2 = 368 Mpa...............(2)
At mirror radius 3 = 0.162 mm Bending stress 3 = 442 Mpa...............(3)
comparing case 1 and 2 using the above equation
Taking the natural logarithm of both side
ln(0.6114) = n ln(0.3366)
n₁ = ln(0.6114)/ln(0.3366)
n₁ = 0.452
comparing case 2 and 3 using the above equation
Taking the natural logarithm of both side
ln(0.8326) = n ln(0.7980)
n₂ = ln(0.8326)/ln(0.7980)
n₂ = 0.821
comparing case 1 and 3 using the above equation
Taking the natural logarithm of both side
ln(0.5090) = n ln(0.2687)
n₃ = ln(0.5090)/ln(0.2687)
n₃ = 0.514
average for n
n = 0.596
Hence to get bending stress x at mirror radius 0.796
stress x = 191 MPa
Answer:
189.15cy
Explanation:
To understand this problem we need to understand as well the form.
It is clear that there is four wall, two short and two long.
The two long are
The two long are
The two shors are
The height and the thickness are 14ft and 0.83ft respectively.
So we only calculate the Quantity of concrete,
That in cubic yards is equal to
Hence, we need order 5% plus that represent with the quantity
Answer:
Rate of heat transfer to the room air per meter of pipe length equals 521.99 W/m
Explanation:
Since it is given that the radiation losses from the pipe are negligible thus the only mode of heat transfer will be by convection.
We know that heat transfer by convection is given by
where,
h = heat transfer coefficient = 10.45 (free convection in air)
A = Surface Area of the pipe
Applying the given values in the above formula we get