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3 years ago

A person who weighs 509,45 N empties her lungs as much as

Physics

1 answer:

Masja [62]3 years ago

7 0

Answer:

The weight of the girl = 1045.86 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³.

From Archimedes principle,

R.d = Density of the person/Density of water = Weight of the person in air/Upthrust.

⇒ D₁/D₂ = W/U............................... Equation 1.

Where D₁ = Density of the person, D₂ = Density of water, W = Weight of the person in air, U = Upthrust in water.

Making D₁ the subject of the equation,

D₁ = D₂(W/U)................................... Equation 2

Given: D₂ = 1000 kg/m³ , W = 509.45 N, U = lost in weight = weight in air - weight in water = 509.45 - 22.34 = 487.11 N

Substituting these values into equation 2

D₁ = 1000(509.45/487.11)

D₁ = 1045.86 kg/m³

Thus the weight of the girl = 1045.86 kg/m³

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3 years ago

A soccer player practices kicking the ball into the goal from halfway down the soccer field. The time it takes for the ball to g

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3 years ago

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Wheat ground into flour is an example of a ___ change?

EleoNora [17]

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3 years ago

A car travels 48 km in 1.2 hours. What is the average speed of the car in km/hr

Alborosie

Answer:

average speed of car is 40 km/h

Explanation:

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5 0

3 years ago

2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th

Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = $5.33$ × $10^{-5}$ T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ $B = \frac{\mu_{0}i }{2\pi R}$

Where $B =$ magnetic field due to long wires, $\mu_{0} =$ $4\pi \times10^{-7}$ , $R =$ perpendicular distance from wire to given point

From any one wire $R_{1} =$ 5 cm, $R_{2} =$ 3 cm

so we write,

∴ $B = B_{1} + B_{2}$

$B = \frac{\mu_{0} i}{2\pi R_{1} } + \frac{\mu_{0} i}{2\pi R_{2} }$

$B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]$

$B = 5.33\times10^{-5} T$

Therefore, the magnitude of magnetic field at given point = $5.33\times10^{-5} T$

3 0

3 years ago