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vodomira [7]
3 years ago
7

A river flows due east at 12 km/h. A boat crosses the 720 m wide river by maintaining a constant velocity of 72 mi/h due north r

elative to the water. If no correction is made for the current, how far downstream does the boat move by the time it reaches the far shore?

Physics
1 answer:
zhenek [66]3 years ago
6 0

Answer:

The boat will be 74 .17 meters downstream by the time it reaches the shore.

Explanation:

Consider the vector diagrams for velocity and distance shown below.

converting 72 miles per hour to km/hr

we have 72 miles per hour 72 × 1.60934 = 115.83 km/hr

The velocity vectors form a right angled triangle, and can be solved using simple trigonometric laws

tan \theta = \frac{12}{115.873}

\theta = tan^{-1}(  \frac{12}{115.873})=5.9126

This is the vector angle with which the ship drifts away with respect to its northward direction.

<em>From the sketch of the displacement vectors,  we can use trigonometric ratios to determine the distance the boat moves downstream.</em>

Let x be the distance  the boat moves downstream.d

sin(5.9126)=\frac{x}{720}

x= 720\times 5.9126

x=74.17m

∴The boat will be 74 .17 meters downstream by the time it reaches the shore.

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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
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A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

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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

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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

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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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