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WARRIOR
1 month 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:
Yuliya22 [3.3K]1 month ago
7 0

The responses are:

a) Work=125,923.61J

b) Power=1.46watt

Rationale:

It appears you neglected to record the questions for the problem, but to assist, I will attempt to complete it.

The questions are:

a) What work does the heart perform in a day?

b) How much power does it generate in watts?

Thus, solving we determine:

We must convert liters to cubic meters to utilize the provided information, hence:

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

Additionally, we need to determine the mass using the blood's density.

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

Now, calculating the work done by the heart in one day, we find:

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

Finally, calculating the output power and its horsepower, we arrive at:

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

Have a nice day!

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Radiation emitted from human skin reaches its peak at λ = 940 µm. (a) What is the frequency of this radiation? (b) What type of
inna [3103]

Answer:

a) The frequency is 3.191x10^11.

b) The type of radiation is microwave radiation.

c) The energy of one quantum of this radiation amounts to 1.32x10^-3ev.

Explanation:

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8 0
2 months ago
1. Use Coulomb’s Law (equation below) to calculate the approximate force felt by an electron at point A in the schematic below.
Keith_Richards [3271]

Answer:

Explanation:

The data indicates that point A is located midway between two charges.

To calculate the electric field at point A, we begin with the field produced by charge -Q ( 6e⁻ ) at A:

= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴

= 13.82 x 10⁻⁶ N/C

This field points towards Q⁻.

A similar field will arise from the charge Q⁺, but it will direct away from Q⁺ toward Q⁻.

To find the resultant field, we add these contributions:

= 2 x 13.82 x 10⁻⁶

= 27.64 x 10⁻⁶ N/C

For the force acting on an electron placed at A:

= charge x field

= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶

= 44.22 x 10⁻²⁵ N

8 0
2 months ago
A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "
kicyunya [3294]

Answer:

a) Q = C*emf

b)  Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let V_{R}represent the Voltage across the Resistor

V_{c}and

represent the Voltage across the capacitor

Implementing KVL;

emf - V_{R} - V_{c} = 0\\

.........................(1)

Since this is a series connection, the same current traverses through the circuit

V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C

Integrating V_{c}and V_{R}into equation (1)

emf - IR - Q/C = 0

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

I = 0A\\emf = Q/C\\Q = C* emf

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\

Substituting the values of t and I₀ into the aforementioned formula for I

I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf

4 0
1 month ago
Which of these nebulae is the odd one out?
Softa [3030]

The correct choice is D!

Clarification:

5 0
1 month ago
Read 2 more answers
An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t
Maru [3345]

Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension present in rope = 550 N

The total torques will be

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7 0
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