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Brilliant_brown [7]
2 years ago
6

The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha

t the work done by the heart is equal to the work required to lift this amount of blood a height equal to that of the average American female (1.63 m). The density (mass per unit volume) of blood is 1.05×103kg/m3.
Physics
1 answer:
gregori [183]2 years ago
7 0

The answers are:

a) Work=125,923.61J

b) Power=1.46watt

Why?

It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.

The questions are:

a) How much work does the heart do in a day?

b) What is its power output in watts?

So, solving we have:

We need to convert from liter to cubic meters in order to use the given information, so:

1L=0.001m^{3}\\\\7500L*\frac{0.001m^{3} }{1L}=7.5m^{3}

Also, we need to find the mass given the density of the blood.

1050}\frac{kg}{m^{3}}*7.5m^{3}=7875kg

Now, calculating how much work does the heart do in a day, we have:

Work=Fd=mgh\\\\Work=7875kg*9.81\frac{m}{s^{2}}*1.63m=125,923.61J

Then, calculating what is the power output and its horsepower, we have:

Power=\frac{Work}{time}\\\\Power=\frac{125,923.61J}{86,400s}=1.46watt

Have a nice day!

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Answer:

<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>

Explanation:

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Since the negative ends are joined together  the total charge on both capacity would be;

q = q_{1} -q_{2}

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In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage

q = (4 x V) + (6 x V)

20 = 10 V

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Therefore the final charge on the 6.0 mF capacitor would be 12 mC

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2 years ago
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A person is working on a steel structure while standing on the ground. An accident occurred where 5 A pass through the structure
matrenka [14]

Answer:

35mA

Explanation:

Hello!

To solve this problem we must use the following steps

1. Find the electrical resistance of the metal rod using the following equation

R=\alpha  \frac{l}{a}

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α=

metal rod resistivity=2x10^-4 Ωm

l=leght=2m

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A=\frac{\pi }{4} d^2=\frac{\pi }{4} (0.06)^2=0.00283

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R=(2x10^-4)\frac{2}{0.00283} =0.14

2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)

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4.as the voltage on both nodes is the same we can use ohm's law in resitance 1 and 2 (V=IR)

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