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Vinvika [58]
3 years ago
14

In a research facility, a person lies on a horizontal platform which floats on a film of air. When the person's heart beats, it

pushes a mass m of blood into the aorta with speed v, and the body and platform move in the opposite direction with speed V. Assume that the blood's speed is 56.5 cm/s. The mass of the person + platform is 54.0 kg. The platform moves 6.30 ✕ 10−5 m in 0.160 s after one heartbeat. Calculate the mass (in g) of blood that leaves the heart. Assume that the mass of blood is negligible compared with the total mass of the person, and the person + platform is initially at rest. (Also assume that the changes in velocity are instantaneous.)
Physics
2 answers:
Nikolay [14]3 years ago
8 0

Answer:

Explanation:

Guven:

Mass of person + platform, Mt = 54 kg

= 54000 g

Velocity, Vb = 56.5 cm/s

Distance, D = 6.3 × 10^-5 m

= 6.3 × 10^-3 cm

Time = 0.16 s

V = distance/time

= 6.3 × 10^-3/0.16

= 0.039375 cm/s

From the question,

Mt × V = Mb × Vb

54000 × 0.039375 = Mb × 56.5

Calculating mass of blood,

Mb = 37.63 g

= 37.63 g of blood

anastassius [24]3 years ago
7 0

Answer: 3.48g

Explanation:

here, we will be using conservation of momentum to solve the problem. i.e the total momentum remains unchanged, unless an external force acts on the system. We'll in thus question, there is no external force acting in the system.

Remember, momentum = mass * velocity, then

mass of blood * velocity of blood = combined mass of subject and pallet * velocity of subject and pallet

Velocity of blood = 56.5cm = 0.565m

mass of blood * 0.565 = 54kg * (0.000063/0.160)

mass of blood * 0.565 = 54 * 0.00039375

mass of blood * 0.565 = 0.001969

mass of blood = 0.00348kg

Thus, the mass of blood that leaves the heart is 3.48g

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Answer:

a) p₂ = 1.88 kg*m/s

   θ = 273.4 º

b)  Kf = 37% of Ko

Explanation:

a)

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  • Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
  • The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
  • So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

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  • We can do the same for the particle moving along the positive y-axis:

        p_{o2x} = 0 (3)

        p_{o2y} = 4.00 kg*m/s (4)

  • Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
  • Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

       p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)

      p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s  (6)

  • Now, the total initial momentum, along these directions, must be equal to the total final momentum.
  • We can write the equation for the x- axis as follows:

       p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x}  (7)

  • We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

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  • Now, we can repeat exactly the same process for the y- axis, as follows:

       p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y}  (9)

  • We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

       p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)

  • Since we have the x- and y- components of the final momentum of  the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

       p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} }  = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)

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  • The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

b)

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       \frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)

  • So, the final kinetic energy has lost a 37% of the initial one.

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