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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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Semenov [28]

Explanation:

Work = force × displacement

532 J = 48 N × d

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3 years ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}.

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3 years ago
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Which undergoes greater acceleration: an airplane that goes from 1000 km/h to 1005 km/h in 10 seconds or a skateboard that
Alex_Xolod [135]

Answer:

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Explanation:

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The airplane's acceleration is:

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roblem 10: In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following p
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Answer:

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p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ

W= \int\limits^f_i {K/V^} \, dV

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