Answer:
5.70 m
Explanation:
3sₐ + sB = L
3∆sₐ = - ∆sB
3vₐ = -vB
3*3 = -vB
vB = -9 ms⁻²
T₁ + V₁ = T₂ + V₂
(0 + 0) + (0 + 0) = 1/2(20)(3)² + 1/2(30)(-9)² + 20*9.81*(sB/3) + 30*9.81* sB
0 = 90 + 1215 - 228.9sB
-1305 = - 228.9sB
sB = 1305/228.9 = 5.70 m
Answer:
the nation will suffer terrible consequences
Explanation:
I did that and got it right
Answer:
As the temperature of a thermal radiator is increased
Group of answer choices
- the object appears redder.
- the object appears bluer.
- the object emits more power for the same area.
- the object emits less power for the same area.
- the object expands to keep the same power per units area.
<em>When the temperature of a thermal radiator increases ;</em>
- <em>the object emits more power for the same are</em>
- <em>the object becomes bluer</em>
Explanation:
Thermal radiation involves the transfer of heat between molecules of two substances without direct contact with each other. When a body is heated to a given temperature it begins to emit light which is transferred to nearby objects as thermal radiation. The medium through which the heat is transferred could be liquid, solid, or in a vacuum.
<h3>How temperature affects thermal radiation.</h3>
Temperature determines the amount of heat that is been radiated from a body. An increase in temperature would increase the thermal radiation of the body. The increase in the heat radiation results to increase in the thermal energy of the body. Also when a body is heated it tends to be bluer than a cool object, this is caused by the rapid movement of the molecules.
Therefore When the temperature of a thermal radiator increases ;
- the object emits more power for the same are
- the object becomes bluer
Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
