The correct answer would be A "<span>A light-year is the distance light travels in a year.
This is considered a unit of distance connected to the distance that light can travel in one year. It is proved that light travels at 300,000 km per second so, in 1 year, it might travel 10 trillion km.
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Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
Answer:
D) The capacitor will pull the material into the space between the plates, and the potential energy stored between the plates of the capacitor will decrease by a factor of 3
Explanation:
points to note
A) inserting a material with dielectric constant of 3 between the plates means that the capacitance C will increase by 3.
B) Since the battery is disconnected, the potential difference V between plate will not remain constant.
C) increasing the capacitance reduces the potential difference across the plate, and from C = Q/V it can be seen that the charges on the plate remains constant.
From the proof in the image below, the reduction in the potential energy of the capacitor is due to the energy used by the capacitor to pull the dielectric material into the space.
The correct answer is "Humidity". good luck!
This question is incomplete. The complete question is given below:
Question 3 Both the angle and the magnitude of the force have a certain uncertainty: εF = 28 N and εθ = 0.8°. Using the propagation methods described in the video you watched at the beginning of this prelab, calculate the corresponding propagated uncertainty for Fx, in N. For this question, round up your final answer to two significant figures. Do not include the ± sign in your answer. Example: If the x component of F is 200±14 N, you should enter “14”.
Both the force and the angle are measured, and the results are quoted as a central value plus/minus an uncertainty:
F = F0 ± εF
θ = θ0 ± εθ
We would like to evaluate the component of the force in the x direction.
Question 2
Let us first concentrate on the central value. Take F0 = 325 N and θ0 = 57°.
The answer & explanation for this question is given in the attachment below.