"Maritime" air . . . . . always high relative humidity
"Polar" air . . . . . always cold
"Maritime polar" air . . . . . choice-A
Answer:
S = V0 t + 1/2 a t^2
S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)
S = 1500 m + .6 * 90000 m = 55,500 m
Check: V0 = 5 m/s
V2 = V0 + a t = 5 + 1.2 * 300 = 365 m/s
Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s (note uniform motion)
S = 185 * 300 = 55,500 m
We calculated V2 above at 365 m/s the speed after 300 sec
Answer:
This will require 266.9 of heat energy.
Explanation:
To calculate the energy required to raise the temperature of any given substance, here's what you require:The mass of the material, m The temperature change that occurs, ΔT The specific heat capacity of the material,
c
(which you can look up). This is the amount of heat required to raise 1 gram of that substance by 1°C.
Here is a source of values of
c for different substances:
Once you have all that, this is the equation:
Q=m×c×ΔT(Q is usually used to symbolize that heat required in a case like this.)For water, the value of c is 4.186g°C So, Q=750×4.186×85=266=858=266.858