Answer
given,
discharge rate from pipe = 1000 gallons/minutes
now,
flow rate in cubic meters per second
1 gallon = 0.00378541 m³
1 min = 60 s
Q =
Q = 0.063 m³/s
flow rate in liters per minute
1 gallon = 3.78541 L
Q =
Q = 3785.41 m³/min
flow rate in cubic feet per second
1 gallon = 0.133681 ft³
1 min = 60 s
Q =
Q = 2.23 ft³/s
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
To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.
The drift velocity is given by the equation:
Where
I = current
n = Number of free electrons
A = Cross-Section Area
q = charge of proton
Our values are given by,
The hall voltage is given by
Where
B= Magnetic field
n = number of free electrons
d = distance
e = charge of electron
Then using the formula and replacing,