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2 years ago

What buoyant force is exerted on a 6,000-mL toy balloon by the air surrounding the balloon? The density of air is 0.0009 g/mL

Physics

1 answer:

Vlada [557]2 years ago

6 0

The buoyant force exerted on a 6,000-mL toy balloon by the air surrounding the balloon is 52.97N.

<h3>How to calculate buoyant force?</h3>

The buoyancy needed for an object can be calculated using the formula;

B = ρ × V × g

where;

ρ and V are the object's density and volume respectively
g is the acceleration due to gravity (9.81m/s²)

B = 6000 × 0.0009 × 9.81

B = 52.97N

Therefore, the buoyant force exerted on a 6,000-mL toy balloon by the air surrounding the balloon is 52.97N.

Learn more about buoyant force at: brainly.com/question/21990136

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Answer:

$Q_a=330 cal$

$Q_w=6000cal$

Explanation:

From the question we are told that

Mass of the aluminum container 50 g

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Initial temperature of the container and water 20°C

Temperature of the steam 100°C

Final temperature of the container, water, and condensed steam 50°C

Mass of the container, water, and condensed steam 261 g

Mass of the steam 11 g Specific heat of aluminum 0.22 cal/g°C

a) Heat energy on container

Generally the formula for mathematically solving heat gain

$Q_c=M_c *C_c*( \triangle T)$

Therefore imputing variables we have

$Q_a=50g *0.22*50-20$

$Q_a=330 cal$

b) Heat energy on water

Generally the formula for mathematically solving heat gain

$Q_w=M_w *C_w*( \triangle T)$

Therefore imputing variables we have

$Q_w=200 *1* 50-204$

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3 years ago

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2 years ago

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Answer:

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This is a fluid mechanics exercise

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area is

A₂ = π/4 d₂²

d₂ = $\sqrt{4 A_2 / \pi }$

d₂ = $\sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }$

d₂ = 10.14 10⁻³ m

d₂ = 1.014 cm

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