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natita [175]
2 years ago
6

An electric power plant uses energy from burning coal to generate steam at 450∘C. The plant is cooled by 20∘C water from a nearb

y river. If burning coal provides 100 MJ of heat, what is the theoretical minimum amount of heat that must be transferred to the river during the conversion of heat to electric energy? A. 100 MJ. B. 90 MJ. C. 60 MJ. D. 40 MJ.
Physics
1 answer:
Goryan [66]2 years ago
6 0

Answer:

40 MJ (D)

Explanation:

Quantity of heat (Qh) = 100 MJ

temperature of steam (Th) = 450°c = 450 + 273 = 723 K

emperature of water (TI) = 20 °c = 20 + 273 = 293 k

efficiency = (Qh-Qi)/Qh = (Th-Ti)/Th

 \frac{100x10x^{6}-Qi }{100x10^{6}} = \frac{723-293}{723}

100x10^{6} - Qi= 0.5947 x 100 x 10 ^{6}

100x10^{6} - (0.5947 x 100x10^{6}) = Qi

Qi = 40.5 MJ equivalent to 40 MJ (D)

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Kaylis [27]

Answer:8.3m/sec 30 sec,

Explanation:

A student practicing for a track meet, ran 250 m in 30 sec. a. What was her average speed? 250 m = 8.3 m/sec 30 sec.

5 0
1 year ago
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What is the concentration of molecular oxygen (O2) in mol/L on a June day in Toronto when atmospheric pressure is 1.0 atm and th
saveliy_v [14]

Answer:

The concentration of mole evil at oxygen on that day is 0.00858 mol/L

Explanation:

Here, we want to calculate the concentration of molecular oxygen

The pressure on that day is 1.0 atm

Since oxygen is at a concentration of 21%, the pressure of oxygen will be 21/100 * 1 = 0.21 atm

Now let’s calculate the concentration;

From Ideal gas law;

PV = nRT

This can be written as;

P/RT = n/V

The term n/V refers to concentration;

Let’s make substitutions now;

P = pressure = 0.21 atm

R = molar gas constant = 0.0821 L•atm/mol•k

T = temperature = 25 = 25 + 273.15 = 298.15 K

Substituting these values, we have;

n/V = C = 0.21/(0.0821 * 298.15) = 0.00858 mol/L

6 0
2 years ago
A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of
GREYUIT [131]

Answer:

a=190\ m/s^2

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2

So, the acceleration of the hockey puck is 190\ m/s^2.

5 0
2 years ago
An American manufacturer supplied a customer with refrigerators with electrical cords that were one yard long instead of 1 meter
Simora [160]

Answer:

C.

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A meter is 8.56 centimeters longer than a yard. Something to keep in mind is that a meter is about 10% longer than a yard.

Hope this helps :)

6 0
1 year ago
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Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a
garik1379 [7]

Answer:

T=2.94*10^-10  N/m.

Explanation:

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20μm, and spider silk has a density of 1300 kg/m³.

To have a fundamental frequency at 150Hz , to what tension must a spider adjust a 14cm -long strand of silk?

l=length of the spider silk, 14cm

velocity of wave = √(T/μ)          

where T = tension and

μ = mass per unit length)

λ/2=l

for fundamental frequency λ/2 =14cm    

 (λ= wavelength of standing wave;  as there will be no node

   except the endpoints of silk strand)

               λ = 28 cm = 0.28 m

and since frequency * wavelength = speed of wave. we have,

                  150 * 0.28 = √(T/μ)                                        ..................(#)

now μ = mass/length = [volume * density]/length = [(length*area) * density] / length = area * density

         = [π * (10 * 10^(-6))²] * 1300  = 13π * 10^(-8).

now putting this in equation (#) we get

    150 * 0.28 = √(T/[13π * 10^(-8)]).

thus T = [13π * 10^(-8)] * (42)²     =  

2.94*10^-10  N/m.

6 0
1 year ago
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