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frutty [35]
3 years ago
5

How would the body be affected if red blood cells and low levels of homoglobinHow would the

Physics
1 answer:
JulsSmile [24]3 years ago
6 0
Low blood pressure. The person could faint and have an irregular heartbeat.
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A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
an elevator mass of 7700 kg falls from a height of 32 m after a sudden failure in the hoisting cable. The mass is stopped by a s
valkas [14]

Answer:k=28.29 kN/m

Explanation:

Given

mass m =7700 kg

height from which Elevator falls h=32 m

Let x be the compression in the spring

thus From conservation of Energy Potential energy will convert in to Elastic Potential Energy of spring

\frac{kx^2}{2}=mg(h+x)----------1

also maximum acceleration is 5g

thus

mg-kx=ma

here a=-5g

kx=mg-m(-5g)=6mg

x=\frac{6mg}{k}

Substitute x in equation 1

0.5\times k\times (\frac{6mg}{k})^2=mg(h+\frac{6mg}{k})

18\frac{(mg)^2}{k}=mgh+6\frac{(mg)^2}{k}

k=12\cdot \frac{mg}{h}

k=12\times \frac{7700\times 9.8}{32}

k=28.29 kN/m

4 0
2 years ago
Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez
katrin [286]

Answer:

a= 4.4×10 m/s^2

Explanation:

pressure P  = E/c

Where, E = 100 W/m^2 intensity of light

c= speed of light  = 3×10^8 m/s

P = 1000/ 3×10^8

P = 3.33×10^(-6) Pa

Force F = P×A

  • P is the pressure and c= speed of light

F = 3.33×10^{-6}×6.65×10(-29)

= 2.22×10^{-6}

acceleration a  = F/m = 2.22×10^{-6}/ 5.10×10^{-27}

a= 4.4×10 m/s^2

4 0
3 years ago
Which type of surface would most likely be the best reflector of electromagnetic energy?
Len [333]
<span>light colored and smooth surface would most likely be the best reflector of electromagnetic energy.Light, shiny surfaces are the best reflectors of radiation and they will allow the waves to reflect and bounce off rather than absorb. we can consider mirror as the example ,it will only reflect the light energy falling on them and it will not absorb. The darker coloured and rough surfaced substances will definitely absorb some amount of light falling on it. so light coloured smooth or shiny surfaced material would be the best reflector for electromagnetic energy.</span>
5 0
2 years ago
An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of
slega [8]

Answer:

T_{1}=94.9^{o}C

Explanation:

Given data

length=100mm

Diameter=5mm

Thermal conductivity=5 W/m.K

Power=50 W

Temperature=25°C

The temperature of heater surface follows from the rate equation written as:

T_{1}=T_{2}+\frac{q}{kS}

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

S=\frac{2\pi L}{ln(\frac{4L}{D} )} \\

Substitute the given values

S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m

The temperature of heater is then:

T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

                           T_{1}=94.9^{o}C

5 0
2 years ago
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