Answer:
The tropospheric tabulation continues to 11,000 meters (36,089 ft), where the temperature has fallen to −56.5 °C (−69.7 °F), the pressure to 22,632 pascals (3.2825 psi), and the density to 0.3639 kilograms per cubic meter (0.02272 lb/cu ft). Between 11 km and 20 km, the temperature remains constant
Explanation:
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Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around
Answer:
660 J/kg/°C
Explanation:
Heat lost by metal = heat gained by water
-m₁C₁ΔT₁ = m₂C₂ΔT₂
-(0.45 kg) C₁ (21°C − 80°C) = (0.70 kg) (4200 J/kg/°C) (21°C − 15°C)
C₁ = 660 J/kg/°C
<span> </span>For any prism-shaped geometry, the volume
(V) is assumed by the product of cross-sectional area (A) and height (h).
<span> V = Ah </span>
<span>
Distinguishing with respect to time gives the
relationship between the rates.
dV/dt = A*dh/dt</span>
<span> in the meantime the area is not altering </span>
<span>
dV/dt = π*(1 ft)^2*(-0.5 ft/min) </span>
<span>
dV/dt = -π/2 ft^3/min ≈ -1.571 ft^3/min
Water is draining from the tank at the rate of π/2
ft^3/min.</span>
