Technician A says that a circuit with continuity reads 0 ohms. Technician B says that an open circuit reads 0 ohms. Who is corre
1 answer:
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Answer:
the crown is false densty= 12556kg/m^3[/tex]
Explanation:
Hello! The first step to solve this problem is to find the mass of the crown, this is found using the weight of the crown in the air by means of the equation for the weight.
W=mg
W=weight(N)=31.4N
M=Mass
g=gravity=9.81m/S^2
solving for M
m=W/g
The second step is find the volume of crown remembering that when an object is weighed in the water the result is the subtraction between the weight of the object and the buoyant force of the water which is the product of the volume of the crown by gravity by density of water
Where
F=weight in water=28.9N
m=mass of crown=3.2kg
g=gravity=9.81m/S^2
α=density of water=1000kg/m^3
V= crown´s volume
solving for V
finally, we remember that the density is equal to the index between mass and volume
To determine the density of the crown without using the weight in the water and with a bucket we can use the following steps.
1.weigh the crown in the air and find the mass
2. put water in a cylindrical bucket and measure its height with a ruler.
3. Put the crown in the bucket and measure the new water level with a ruler.
4. Subtract the heights, and find the volume of a cylinder knowing the difference in heights and the diameter of the bucket, in order to determine the volume of the crown.
5. find density by dividing mass by volume
Answer:
Ts =Ta E)- 300(
569.5 K
5
Q12-W12 = -4014.26
Mol
AU2s = Q23= 5601.55
Mol
AUs¡ = Ws¡ = -5601.55
Explanation:
A clear details for the question is also attached.
(b) The P,V and T for state 1,2 and 3
P =1 bar Ti = 300 K and Vi from ideal gas Vi=
10
24.9x10 m
=
P-5 bar
Due to step 12 is isothermal: T1 = T2= 300 K and
VVi24.9 x 10x-4.9 x 10-3 *
The values at 3 calclated by Uing step 3l Adiabatic process
B-P ()
Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=
14
Ps-1x(4.99 x 103
P-1x(29x 10)
9.49 barr
And Ts =Ta E)- 300(
569.5 K
5
(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and
Work done for Isotermal process define as
8.314 x 300 In =4014.26
Wi2= RTi ln
mol
And fromn first law of thermodynamic
AU12= W12 +Q12
Q12-W12 = -4014.26
Mol
F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and
Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-
AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53
Inol
Now from first law of thermodynamic the Q23
AU2s = Q23= 5601.55
Mol
For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0
and
AH
C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18
mol
AU=C, (T¡-T)= x 8.314 (300
-5601.55
569.5)
mol
Now from first law of thermodynamie the Ws1
J
mol
AUs¡ = Ws¡ = -5601.55
