Answer:
O is truse is the best answer hhahahha
Explanation:
Given Information:
Initial temperature of aluminum block = 26.5°C
Heat flux = 4000 w/m²
Time = 2112 seconds
Time = 30 minutes = 30*60 = 1800 seconds
Required Information:
Rise in surface temperature = ?
Answer:
Rise in surface temperature = 8.6 °C after 2112 seconds
Rise in surface temperature = 8 °C after 30 minutes
Explanation:
The surface temperature of the aluminum block is given by
Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.
After t = 2112 sec:
The rise in the surface temperature is
Rise = 35.1 - 26.5 = 8.6 °C
Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.
After t = 30 mins:
The rise in the surface temperature is
Rise = 34.5 - 26.5 = 8 °C
Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.
Answer:
b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.
Explanation:
From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.
The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.
However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.
The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.
It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.
As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
