Answer:
The correct answer is "1341.288 W/m".
Explanation:
Given that:
T₁ = 300 K
T₂ = 500 K
Diameter,
d = 0.2 m
Length,
l = 1 m
As we know,
The shape factor will be:
⇒ ![SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}](https://tex.z-dn.net/?f=SF%3D%5Cfrac%7B2%20%5Cpi%20l%7D%7Bln%5B%5Cfrac%7B1.08%20b%20%7D%7Bd%7D%20%5D%7D)
By putting the value, we get
⇒ ![=\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%20%5Cpi%20l%7D%7Bln%5B%5Cfrac%7B1.08%5Ctimes%201%7D%7B0.2%7D%20%5D%7D)
⇒ 
hence,
The heat loss will be:
⇒ 
Answer:
Answer: ±0.02 units or 20±0.02 units or 19.98-20.02 units depending on how they prefer its written (typically the first or second one)
Explanation:
says on the sheet. Unless otherwise stated 0.XX = ±0.02 tolerance
(based on image sent in other post)
Answer:
T = 5416.67 N
T = -2083.5 N
T = 0
Explanation:
Forward thrust has positive values and reverse thrust has negative values.
part a
Flight speed u = ( 150 km / h ) / 3.6 = 41.67 km / s
The thrust force represents the horizontal or x-component of momentum equation:
Answer: The thrust force T = 5416.67 N
part b
Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:
Answer: The thrust force T = -2083.5 N reverse direction
part c
Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero as there is no difference in two velocities in x direction.
Answer: T = 0 N
Answer:
Springs store energy when compressed and release energy when they rebound
Explanation:
Answer:
(i) 12 V in series with 18 Ω.
(ii) 0.4 A; 1.92 W
(iii) 1,152 J
(iv) 18Ω — maximum power transfer theorem
Explanation:
<h3>(i)</h3>
As seen by the load, the equivalent source impedance is ...
10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω
The open-circuit voltage seen by the load is ...
(36 V)(12/(24 +12)) = 12 V
The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.
__
<h3>(ii)</h3>
The load current is ...
(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current
The load power is ...
P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power
__
<h3>(iii)</h3>
10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...
(600 s)(1.92 J/s) = 1,152 J
__
<h3>(iv)</h3>
The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.
The power transferred to 18 Ω is ...
((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W
