Correct answer choice is:
D. A continuous transmission of energy from one location to the next.
Explanation:
Waves include the carrier of energy without the carrier of matter. In outcome, a wave can be characterized as a change that progresses into a medium, carrying energy from one spot (its source) to different spot without carrying matter.
The parietal layers of the membranes line the walls of the body cavity.
hope this helps
The solution for the problem is:
Constant speed means Fnet = 0.
Let m = mass of wood block and Θ = angle of ramp; then if µk = 0.35 …
The computation would be:
Fnet = 0 = mg (sin Θ) - (µk) (mg) (cos Θ)
mg (sin Θ) = µk (mg) (cos Θ)
µk = tan Θ
Θ = arctan(µk)
= arctan (0.35)
≈ 19.3°
I think the correct answer from the choices listed above is option A. The rent is an<span> example of a monthly fixed cost for a sandwich shop. It is a fixed cost since you are required to pay for it per month. Hope this answers the question. Have a nice day.</span>
Answer:
change in internal energy 3.62*10^5 J kg^{-1}
change in enthalapy 5.07*10^5 J kg^{-1}
change in entropy 382.79 J kg^{-1} K^{-1}
Explanation:
adiabatic constant 
specific heat is given as 
gas constant =287 J⋅kg−1⋅K−1

specific heat at constant volume

change in internal energy 

change in enthalapy 

change in entropy


