Where r is the radius of balloon.
Here mass of woman = 68 kg
Mass of air displaced by a balloon with volume V = 1.29*V
Mass of helium inside balloon = 0.178*V
Total mass to be lifted by balloon = 68 +0.178*V
Buoyant force = 1.29V-0.178V=1.112V
So we have 1.112 V = 68+ 0.178*V
0.934 V = 68
V = 72.81 
\frac{4}{3} \pi r^{3}[/tex]= 72.81
r = 2.59 m
So radius of helium balloon = 2.59 m
Because the temperature remains constant, we can apply Boyle's Law which states that
pV = constant
where
p = pressure
V = volume
Define the two states of the gas.
State 1
Pressure = p₁
Volume = 1000 ml
State 2
Pressure = p₂
Volume = 500 ml
Apply Boyle's law.
1000p₁ = 500p₂
2 = p₂/p₁
By halving the volume, the pressure doubles.
Answer:
The pressure increases by a factor of 2.
Answer:
The answer is C.
120 V with 60 W light bulb is 240 ohms.
120 V with 100 W light bulb is 144 ohms.
The 100 W bulb has less resistance :)
Add the KE increase and the work done against friction.
The final velocity is twice the average, or 3.0 m/s
The final KE is (1/2)*25*3^2 = 112.5 J
The friction work done is 6*3.8 = 22.8 J
hope this is correct
Adding thermal energy
Performing work on the system
