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LenKa [72]
3 years ago
10

Experiments need to be repeated by the scientist and also replicated by other scientists.

Physics
2 answers:
BlackZzzverrR [31]3 years ago
6 0

Answer:

true

Explanation:

grandymaker [24]3 years ago
3 0

Answer:

false

Explanation:  because the other scientists could've missed something in there experiment

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A monatomic ideal gas has pressure p1 and temperature T1. It is contained in a cylinder of volume V1 with a movable piston, so t
Vanyuwa [196]

Answer:

A) Q1 = (3/2)P1V1[A - 1]

B) W2 = P1V1(In A)

C) W3 = P1V1(1 - A)

Explanation:

A) From first law of thermodynamics and applying to the question, we have;

ΔU = Q - W

Where,

ΔU = change in internal energy

Q = the heat absorbed

W = the work done

Now, because the first process occurs at constant volume, the work done is zero:

Thus,

ΔU = Q - 0

ΔU = Q

The change in internal energy is given by;

ΔU = nCvΔt

where;

n = the number of moles of the gas

R = the gas constant,

Cv = the specific heat at constant volume

Δt = The change in temperature i.e T2 - T1.

Now, using the ideal gas law, let us find an expression for n and Δt

P1V1 = nRT1

n = P1V1/RT1

T1 = P1V1/nR

Now, the specific heat at constant volume is Cv = (3/2)R

Now, from the question, since it's pressure has reached AP1, we can calculate the temperature T2 by using the ideal gas law at the new conditions of the gas as;

AP1V1 = nRT2

T2 = AP1 V1/ nR

Now, we are to express the heat added in terms of p1, V1, and A

Q = ΔU = nCv(T2 - T1)

From earlier, we saw that,

T1 = P1V1/nR

Putting equation of T2 and T1 into the energy equation to get;

Q = nCv((AP1 V1/ nR) - P1V1/nR)

Q = Cv • P1V1/R (A - 1)

Now, from earlier, we saw that Cv = (3/2)R. Thus,

Q = (3/2)R • P1V1/R (A - 1)

Q = (3/2)P1V1[A - 1]

B) Here again, we are to express work done in step 2 in terms of p1, V1, and A.

This process is an isothermal process because temperature is constant and so work done is given as; W = nRT In(V2/V1)

T = T1 because temperature is constant

From earlier, we saw that;

n = P1V1/RT1 and

But in this process, it's

n = P1V1/RT1 and thus,

V2 = nRT2/P1

We also saw that T2 = AP1 V1/ nR

V1 = nRT2/AP1

Plugging in the relevant values into, W = nRT In(V2/V1), we obtain;

W = (P1V1/RT1) • RT1 • In((nRT2/P1)/(nRT2/AP1)

W = P1V1(In A)

C) In step 3,we have and isobaric process because the pressure is constant.

Work done in this case is given by ;

W = P(V1 - V2)

Because V2 in now the final volume while V1 is now the the initial volume

Now, P is P1 because it's an isobaric process.

From earlier, we saw that,

V1 = nRT2/AP1 and V2 = nRT2/P1

And that T2 = AP1 V1/ nR

Thus,

V1 = V1 and V2 = AV1

Thus, W = P1(V1 - AV1) = P1V1(1 - A)

4 0
3 years ago
A piston–cylinder assembly fitted with a slowly rotating paddle wheel contains 0.13 kg of air, initially at 300 K. The air under
olganol [36]

Answer:

Explanation:

From the first law of thermodynamics

Q=ΔU+W

Q=heat supplies to the system

ΔU=change in internal energy of the system

W= work done by the gas  by the system

Q =12 KJ

ΔU=U_{2} -U_{1}=(mR/k-1)ΔT

ΔU=\frac{0.3*0.287}{1.4-1}*(460-400)

     =14.924 KJ

Q=ΔU+W

W=-2.924 KJ

W_{net}=W_{air}+W_{paddle}

W_{air}=work done by the air

W_{air}= P(V2-V1)

W_{air}=mRT2-mRT1

W_{air}=mR(T2-T1)=0.138*0.287*(460-300)=

                           =6.33 KJ

W_{net}=W_{air}+W_{paddle}

                      =-2.924-6.33

W_{paddle}=-9.26  KJ                    

7 0
3 years ago
Consider the force field and circle defined below. F(x, y) = x2 i + xy j x2 + y2 = 121 (a) Find the work done by the force field
kirza4 [7]

Answer: the work done by the force is 0

Explanation:

F (x², xy)

121 = 11²

so R = x² + y² = 11²

p = x². Q = xy

Δp/Δy = 0, ΔQ/Δx

using Green's theorem

woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA

=  (x² + y² = 121)_∫∫ yΔA

now let x = rcosФ, y = rsinФ

ΔA = rΔrΔФ

so r from 0 to 11

and Ф from 0 to 2π

= 0_∫^2π   0_∫^11  rsinФ × rΔrΔФ

= 0_∫^2π SinФΔФ   0_∫^11  r²Δr

= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0

therefore the work done by the force is 0

3 0
4 years ago
PLEASE HELP
Ronch [10]
Seems the both of us haven't done the project yet.
6 0
3 years ago
Anders suffered a shock when his electric radio dropped into the bathtub--while anders was taking a bath. Anders argued that he
blsea [12.9K]

The claim Anders is most likely to make is the failure of the manufacturer to warn about such risk.

<h3>What is a Risk?</h3>

This is defined as the possibility of something bad happening and in this case it is electric shock when dropped into the bathtub.

Anders can decide to sue for not warning against risk of electric shock when in contact with water.

Read more about Risk here brainly.com/question/1224221

3 0
2 years ago
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