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Andrew [12]
3 years ago
7

A rigid tank contains air at a pressure of 70 psia and a temperature of 55 ˚F. By how much will the pressure increase as the tem

perature is increased to 115 ˚F?
Physics
1 answer:
Katyanochek1 [597]3 years ago
7 0

Answer:

P_2 = 62.69 psi

Explanation:

given,

P₁ = 70 psia               T₁ = 55° F    = (55 + 459.67) R

P₂ = ?                         T₂ = 115° F    = (115 + 459.67) R

we know,

p = ρ RT

ρ is the density which is constant

R is also constant

now,

\dfrac{P_2}{P_1} =\dfrac{T_2}{T_1}

\dfrac{P_2}{70} =\dfrac{55+459.67}{115+459.67}

    P_2 = 62.69 psi

Hence, the increase in Pressure is equal to P_2 = 62.69 psi

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A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o
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Answer:

Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ

Explanation:

For this case we know the mass of the water given :

m = 787 gr

And we know that the initial temperature for this water is T_i =75 C.

We want to cool this water to the human body temperature T_f = 37 C

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

Q= m c_p \Delta T

Where c_p represent the specific heat for the water and this value from tables we know that c_p =1 \frac{cal}{gr C} for the water.

So then we have everything in order to replace into the formula of sensible heat and we got:

Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ

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3 years ago
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what is the resistance of a car light bulb that conducts 0.025A current when connected to a 12V car accumulator? is the current
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One form of Ohm's Law says . . . . . Resistance = Voltage / Current .

R = V / I

R = (12 v) / (0.025 A)

R = (12 / 0.025) (V/I)

<em>R = 480 Ohms</em>

I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is.  (Floogle isn't sure either.)

If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.

If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.  

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3 years ago
Light incident on a lake surface is partly reflected and partly refracted.What is the differences between the reflected ray and
iren2701 [21]

Answer: As per the question, a ray of light is incident on a surface and it is partly reflected and refracted. The incident light is an unpolarised light. The reflected light is partially polarised.

If the angle of incidence becomes equal to the Brester angle (polarising angle), then the reflected light is completely plane polarised.

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A measure of randomness or disorder
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Rx: hydrochloric acid 2 percent solution 500 ml. your stock solution of hydrochloric acid is 10 percent. how much of the stock s
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100 ml

100 ml of the stock solution is required to prepare the order.

We know that C1V1 = C2V2

where C1= 2%

V1 = 500ml

C2= 10%

V2 = ?

V2 = C1V1 / C2

     = 500 * 2% / 10%

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V2 = 100 ml

<h3>What is meant by stock solution?</h3>
  • A stock solution is a sizable amount of a typical reagent in a standardized concentration, like sodium hydroxide or hydrochloric acid.
  • This phrase is frequently used in analytical chemistry while doing operations like titrations where it's crucial to employ precise solution concentrations.
<h3>What distinguishes a standard solution from a stock solution?</h3>
  • The main distinction between stock solution and standard solution is that the former is a highly concentrated solution while the later is a solution whose concentration is precisely known.
  • Because standard solutions frequently arrive as stock solutions, the phrases "stock solution" and "standard solution" are connected.

To learn more about stock solution preparation visit:

brainly.com/question/14667249

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