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Diano4ka-milaya [45]
4 years ago
13

The average human has a density of 945 kg/m3 after inhaling and 1020 kg/m3 after exhaling. (a) Without making any swimming movem

ents, what percentage of the human body would be above the surface in the Dead Sea (a body of water with a density of about 1230 kg/m3) in each of these cases?
Physics
1 answer:
Gala2k [10]4 years ago
8 0

Answer:

20% of the human body float above the surface in the dead sea.

Explanation:

Given data,

The human density after inhaling, d₁ = 945 kg/m³

The human density after exhaling, d₂ = 1020 kg/m³

Therefore the average human density, d = 982.5 kg/m³

The density of the dead sea, D = 1230 kg/m³

The percentage of density of human body to the dead sea is,

                                   P % = (982.5 / 1230) x 100 %

                                           = 80 %

Therefore, the human body has 80% of dead sea density.

Hence, 20% of the human body float above the surface in the dead sea.

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A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average forc
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Explanation:

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Impulse=F_{avg}\cdot t=change\ in\ momentum

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Initial Momentum(P_i)=m\times 0

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The difference between the measures of the intercepted arcs of the given circle is 45°.

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We are given that the angle subtended by secant and tangent, outside the circle is,

\theta =90^{\circ}

And angle measured inside the circle is,

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\theta' = \theta - \phi\\\\\theta' = 90-45\\\\\theta' =45^{\circ}

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3 years ago
A ball dropped from a bridge takes three seconds to reach the water below how far is the bridge above the water?
tatuchka [14]

<u>Given that:</u>

Ball dropped from a bridge at the rate of 3 seconds

Determine the height of fall (S) = ?

      As we know that, S = ut + 1/2 ×a.t²

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                          a= g =9.81 m/s  (since free fall)

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