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ArbitrLikvidat [17]
3 years ago
9

A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above w

ater when his lungs are full.
Required:
a. Calculate the volume of air he inhales - called his lung capacity - in liters.
b. Does this lung volume seem reasonable?
Physics
1 answer:
Dima020 [189]3 years ago
4 0

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

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What total energy (in J) is stored in the capacitors in the figure below (C1 = 0.900 µF, C2 = 16.0 µF) if 1.80 10-4 J is stored
Musya8 [376]

The total energy  stored in the capacitors is determined as  2.41 x 10⁻⁴ J.

<h3>What is the potential difference of the circuit?</h3>

The potential difference of the circuit is calculated as follows;

U = ¹/₂CV²

where;

  • C is capacitance of the capacitor
  • V is the potential difference

For a parallel circuit the voltage in the circuit is always the same.

The energy stored in 2.5 μf capacitor is known, hence the potential difference of the circuit is calculated as follows;

U = ¹/₂CV²

2U = CV²

V = √2U/C

V = √(2 x 1.8 x 10⁻⁴ / 2.5 x 10⁻⁶)

V = 12 V

The equivalent capacitance of C1 and C2 is calculated as follows;

1/C = 1/C₁ + 1/C₂

1/C = (1)/(0.9 x 10⁻⁶)  +  (1)/(16 x 10⁻⁶)

1/C = 1,173,611.11

C = 1/1,173,611.11

C = 8.52 x 10⁻⁷ C

The total capacitance of the circuit is calculated as follows;

Ct = 8.52 x 10⁻⁷ C   +   2.5 x 10⁻⁶ C

Ct = 3.35 x 10⁻⁶ C

The total energy of the circuit is calculated as follows;

U =  ¹/₂CtV²

U =  ¹/₂(3.35 x 10⁻⁶ )(12)²

U = 2.41 x 10⁻⁴ J

Learn more about energy stored in a capacitor here: brainly.com/question/14811408

#SPJ1

7 0
1 year ago
PLEASE HELP, I don’t know how to do this
MrRa [10]

Explanation:

7) Given:

v₀ = 2.0 m/s

v = 0 m/s

t = 3.00 s

Find: Δx

Acceleration isn't included in the problem, so use a kinematic equation that doesn't involve a.

Δx = ½ (v + v₀) t

Δx = ½ (0 m/s + 2.0 m/s) (3.00 s)

Δx = 3.0 m

8) Given:

v₀ = 0 m/s

v = 5 m/s

t = 4 s

Find: a

Displacement isn't included in the problem, so use a kinematic equation that doesn't involve Δx.

v = at + v₀

5 m/s = a (4 s) + 0 m/s

a = 1.25 m/s²

9) Given:

v_avg = Δx / t

0.5 m/s = 8 m / t

t = 16 s

7 0
3 years ago
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