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KengaRu [80]
3 years ago
7

1. A hot air balloon weighing 30 N is tied to the ground by a string to prevent

Physics
1 answer:
Mumz [18]3 years ago
5 0

Answer:

a) FB = 260 [N]

b) FT = 230 [N]

Explanation:

In order to solve this problem we must use a static analysis, since Globe does not move. For a better understanding in solving this problem, a free body diagram with the forces acting on the globe is attached.

The buoyant force acts upward as it causes the balloon to tend to float, the weight of the balloon tends to lower the balloon and the downward tension force does not allow the balloon to float

The buoyant force is defined by the following equation:

FB = Ro*V*g

where:

FB = Buoyant force [N]

Ro = density of the air = 1.3 [kg/m^3]

V = volume of the balloon = 20 [m^3]

g = gravity acceleration = 10 [m/s^2]

FB = 1.3*20*10 = 260 [N]

Now we do a sum of forces equal to zero in the y-axis

FB - 30 - FT = 0

260 - 30 = FT

FT = 230 [N]

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Answer:

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(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

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Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

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r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

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