Answer:
https://earthobservatory.nasa.gov/features/EnergyBalance#:~:text=The%20atmosphere%20absorbs%2023%20percent,surface%20radiates%20only%2012%20percent.
Explanation:
Answer:
Kindly check explanation
Explanation:
Given the data:
Trial V (volts) I (mA)
1 1.00 7.2
2 2.10 14.0
3 3.10 20.7
4 4.00 27.2
5 4.90 32.2
Slope = Rise / Run
Rise = y2 - y1 = 32.2 - 7.2 = 25
Run = x2 - x1 = 4.9 - 1.0 = 3.9
Slope = 25 / 3.9 = 6.410
y = mx + c
The intercept, c
Take the point ( 1; 7.2)
Put x = 0
7.2 = 6.410(1) + C
7.2 - 6.410 = C
C = 0.79
Answer:
T0=390 degF
Explanation:
The thermometer reading 70◦Fis placed in an oven preheated to a constant temperature. Through the glass window in the oven door, an observer records that the thermometer reads 110◦F after 12 minutes and 145◦F after 1 minute. How hot is the oven? Newton’s law of cooling yields the following differential equationdTdt=k(T−T0), whereT0 is the ambient temperature.
If T is temperature, then by Newton’s law
dTdt=-k(T−T0)
Where
T0− is temperature of oven
∫
ln(T-T0), from 110 to 70=-kt, from 1/2 to 0
ln(110-T0)-ln(70-T0)=-k(1/2-0)

integrating the LHS of the equation from 110 to 145F
∫∫∫
ln(145-T0)-ln(110-T0)=-k(1-1/2)
2ln145-T0/(110-T0)=-k
k=-2lnT0-145/(T0-110).......................2
couplijng equatiuon 2 with 1
(T0-110)^2=(T0-70)(T0-145)
T0^2-220T0+12100=T0^2-215T0+10150
5T0=1950
T0=390 degF
B, gossiping is unproessional