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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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Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

Mm=44.0 g/mol is the molar mass of the carbon dioxide

So, the number of moles of the gas is

n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol

Now we can find the pressure of the gas by using the ideal gas equation:

pV=nRT

where

p is the pressure

V=1.00 L = 0.001 m^3 is the volume

n = 0.023 mol is the number of moles

R=8.314 J/mol K is the gas constant

T=25.0^{\circ}+273=298 K is the temperature of the gas

Solving the equation for p, we find

p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa

And since we have

1 atm = 1.01\cdot 10^5 Pa

the pressure in atmospheres is

p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm

5 0
2 years ago
Subatomic particles that do not possess any charge but provide mass to atoms are called
WARRIOR [948]

Answer:

Neutrons

Explanation:

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6 0
2 years ago
Two identical ladders are 3.0 m long and weigh 600 N each. They are connected by a hinge at the top and are held together by a h
ruslelena [56]

Answer:

The tension in the rope is 281.60 N.

Explanation:

Given that,

Length = 3.0 m

Weight = 600 N

Distance = 1.0 m

Angle = 60°

Consider half of the ladder,

let tension be T, normal reaction force at ground be F, vertical reaction at top hinge be Y and horizontal reaction force be X.

Y+F=600....(I)

X=T.....(II)

On taking moment about base

X\times l\cos\theta+Y\times l\sin\theta-F\dfrac{l}{2}\sin\theta-T\times d=0

Put the value into the formula

X\times3\cos30+Y\times3\sin30-600\times1.5\sin30-T\times1=0

3\cos30 T-T=600\times1.5\sin30-Y \times3\sin30

1.598T=450-1.5(600-F)....(III)

We need to calculate the force for ladder

2F=600\trimes  2

F=600\ N

We need to calculate the tension in the rope

From equation (3)

1.598T=450-1.5(600-600)

1.598T=450

T=\dfrac{450}{1.598}

T=281.60\ N

Hence, The tension in the rope is 281.60 N.

7 0
2 years ago
Read 2 more answers
A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol
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Answer:

Part a)

v = \sqrt{v_o^2 + g^2t^2}

Part b)

d = \frac{1}{2}gt^2

Part c)

v_f = v_o - gt

Part d)

d = \frac{1}{2}gt^2

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

v_y = -gt

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{v_o^2 + g^2t^2}

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

d = \frac{1}{2}gt^2

Part c)

If helicopter is rising upwards with uniform speed

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v_f = v_i + at

v_f = v_o - gt

Part d)

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mote1985 [20]

Answer:

E=0

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E=\sigma/\varepsilon

in the region outside the plates, the field due to the plates is in opposite directions

E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=-\sigma/2\varepsilon+\sigma/2\varepsilon

E=0

4 0
2 years ago
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