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BigorU [14]
4 years ago
9

A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the

balloon is 0.18 kg/m3. The radius of the balloon (when filled) is R = 4.5 m. The total mass of the empty balloon and basket is mb = 123 kg and the total volume is Vb = 0.068 m3. Assume the average person that gets into the balloon has a mass mp = 74 kg and volume Vp = 0.077 m3.What is the volume of helium in the balloon when fully inflated?
Physics
1 answer:
max2010maxim [7]4 years ago
3 0

Answer:

V = 381.70 m³

Explanation:

ρ air = 1.28 kg / m³

ρ helium = 0.18 kg / m³

R = 4.5 m

Vb = 0.068 m³

mb = 123 kg

To determine the volume of helium in the balloon when fully inflated

V = 4 / 3 π * R ³

V = 4 * π / 3 ( 4.5 m )³

V = 381.70 m³

To determine the mass total

m = ρ helium * V

m = 0.18 kg / m³  * 381.70  m³

m = 68.70 kg

mt = ( 68.70 + 123 )kg

mt = 191.70 kg

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7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

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3 years ago
Choose the best description of the attraction between you and your classmate.
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Answer:

The correct option is;

D. There is not enough information to answer this question

Explanation:

The universal gravitational constant = 6.67408 × 10⁻¹¹ 3³/(kg·s²)

For an in between distance of 1 m and equal masses of 60 kg, we have;

F = G \times \dfrac{m_{1}  \times m_{2}}{r^{2}} = 6.67408 \times 10^{-11}  \times \dfrac{60 \times 60}{1^{2}}  \approx  2.403 \times 10^{-7} N

The gravitational attraction ≈ 2.403 × 10⁻⁷ N, which does not correspond with the answers, therefore, the best option is that there is not enough information to answer this question.

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