a star that suddenly increases greatly in brightness because of a catastrophic explosion that ejects most of its mass.
Answer:
![(a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A](https://tex.z-dn.net/?f=%28a%29%5C%20F_%7BNo%7D%20%3D%20%5BP_%7BNo%7D%20-%20%5Cfrac%7BP_%7Barea%7D%7D%7B2%7D%5D%2A%20A)

Explanation:
Given
---- Altitude of container in Denver
-- Surface Area of the container lid
--- Air pressure in Denver
--- Air pressure in New Orleans
<em>See comment for complete question</em>
Solving (a): The expression for 
Force is calculated as:

The force in New Orleans is:

Since the inside pressure is half the pressure at sea level, then:

Where
--- Standard Pressure
Recall that:

This gives:
![F_{No} = [P_{No} - \frac{P_{area}}{2}]* A](https://tex.z-dn.net/?f=F_%7BNo%7D%20%3D%20%5BP_%7BNo%7D%20-%20%5Cfrac%7BP_%7Barea%7D%7D%7B2%7D%5D%2A%20A)
Solving (b): The value of 
In (a), we have:
![F_{No} = [P_{No} - \frac{P_{area}}{2}]* A](https://tex.z-dn.net/?f=F_%7BNo%7D%20%3D%20%5BP_%7BNo%7D%20-%20%5Cfrac%7BP_%7Barea%7D%7D%7B2%7D%5D%2A%20A)
Where



So, we have:
![F_{No} = [100250 - \frac{101000}{2}] * 0.0155](https://tex.z-dn.net/?f=F_%7BNo%7D%20%3D%20%5B100250%20-%20%5Cfrac%7B101000%7D%7B2%7D%5D%20%2A%200.0155)
![F_{No} = [100250 - 50500] * 0.0155](https://tex.z-dn.net/?f=F_%7BNo%7D%20%3D%20%5B100250%20-%2050500%5D%20%2A%200.0155)


Answer:
The answer to your question is: D) Ф₂ = 49.71°
Explanation:
Data
n₁ = 1.33
Ф₁ = 35°C
n₂ = 1
Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)
Process
Substitution
Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)
Ф₂ = sin⁻¹ (1.33 sin 35/1)
Ф₂ = sin⁻¹ (1.33 x 0.574/ 1)
Ф₂ = sin⁻¹ ( 0.7628 / 1)
Ф₂ = sin⁻¹ (0.7628)
Ф₂ = 49.71°
The force of Earth's gravity on you is greatest when you stand as
close as you can to the center of the Earth (but not underground).
From the choices listed, that would be (c).
<span>(a).the heat trasfer surface area and heat flux on the surface of filament are
Area of Surface= µDL=3.14(0.05cm)(5cm)= 0.785 cm square
qs=Q/Area of surface= 150W/0.785= 191W/cmsq.=1.91x10Âłx10ÂłW/Msq
(b). the heat surface on the surface of heat bulb
Area of surface = 3.14xD²= 3.14(8CM)²= 201.1cm²
qs=Q/Area of surface=150w/201.1cm²=0.75 w/cm²= 7500w/m²
the amount and cost of electrical energy consumed during one period is
Electrical Consumption=QΛt=(0.15 KW)(365X8h/yr)=438 k Wh/yr
Annual cost= 438 kWh/yr)($.08/kWh)= $ 35.04 /yr</span>