Answer: (a) 1.11 nm/mm
Explanation:
Monochromator is a device which is used to transmits a delectable band of wavelengths of light wavelengths available at the input.
Assuming focal length monochromator equipped with a 1200-groove/mm grating
The formula for the first-order reciprocal linear dispersion is
where F is the focal length and n belongs to order of the first spectra.
Make sure you have used the conversion factors correctly as it will have a major impact on the calculation of the answer.
Answer:
2.77mpa
Explanation:
compressive strength = 20 MPa. We are to find the estimated flexure strength
We calculate the estimated flexural strength R as
R = 0.62√fc
Where fc is the compressive strength and it is in Mpa
When we substitute 20 for gc
Flexure strength is
0.62x√20
= 0.62x4.472
= 2.77Mpa
The estimated flexure strength is therefore 2.77Mpa
Answer:
9.6 V
Explanation:
Total resistance (Rt) = R1 + R2 = 500 + 2000 = 2500 ohm
V = IRt
I = V/Rt = 12/2500 = 4.8×10^-3 A
Voltage across the 2000 ohm resistor (V2) = IR2 = 4.8×10^-3 × 2000 = 9.6 V
The type of type of building construction can include materials with no fire-resistance ratings in limited quantities is type 2.
Although many buildings appear to be similar at first appearance, their cost and durability—particularly in an emergency—are affected by the underlying materials. Building codes categorize all structures into Types 1 through 5, and each Type discloses important details like fire-resistance.
Some contemporary structures are now stronger and less expensive to construct. Engineered wood and synthetic plastics, which burn readily, cause quick collapses and present significant risks for firefighters.
Type 1 constructions, which are the most fire-resistant buildings, are made of shielded steel and concrete because they can sustain high temperatures without collapsing. Type 5 constructions, on the other hand, are the least fire-resistant since they are built of flammable materials, are lightweight, and they burn out quickly. Type 2 constructions, modern structures with metal roofs and tilt-slab or reinforced masonry walls.
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Answer:
See explanation
Explanation:
Given:
Initial pressure,
p
1
=
15
psia
Initial temperature,
T
1
=
80
∘
F
Final temperature,
T
2
=
200
∘
F
Find the gas constant and specific heat for carbon dioxide from the Properties Table of Ideal Gases.
R
=
0.04513
Btu/lbm.R
C
v
=
0.158
Btu/lbm.R
Find the work done during the isobaric process.
w
1
−
2
=
p
(
v
2
−
v
1
)
=
R
(
T
2
−
T
1
)
=
0.04513
(
200
−
80
)
w
1
−
2
=
5.4156
Btu/lbm
Find the change in internal energy during process.
Δ
u
1
−
2
=
C
v
(
T
2
−
T
1
)
=
0.158
(
200
−
80
)
=
18.96
Btu/lbm
Find the heat transfer during the process using the first law of thermodynamics.
q
1
−
2
=
w
1
−
2
+
Δ
u
1
−
2
=
5.4156
+
18.96
q
1
−
2
=
24.38
Btu/lbm
