Answer:
q_poly = 14.55 KJ/kg
Explanation:
Given:
Initial State:
P_i = 550 KPa
T_i = 400 K
Final State:
T_f = 350 K
Constants:
R = 0.189 KJ/kgK
k = 1.289 = c_p / c_v
n = 1.2 (poly-tropic index)
Find:
Determine the heat transfer per kg in the process.
Solution:
-The heat transfer per kg of poly-tropic process is given by the expression:
q_poly = w_poly*(k - n)/(k-1)
- Evaluate w_poly:
w_poly = R*(T_f - T_i)/(1-n)
w_poly = 0.189*(350 - 400)/(1-1.2)
w_poly = 47.25 KJ/kg
-Hence,
q_poly = 47.25*(1.289 - 1.2)/(1.289-1)
q_poly = 14.55 KJ/kg
Answer:
the magnitude of the velocity of one particle relative to the other is 0.9988c
Explanation:
Given the data in the question;
Velocities of the two particles = 0.9520c
Using Lorentz transformation
Let relative velocity be W, so
v
= ( u + v ) / ( 1 + ( uv / c²) )
since each particle travels with the same speed,
u = v
so
v = ( u + u ) / ( 1 + ( u×u / c²) )
v = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )
we substitute
v = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )
v = 1.904c / ( 1 + 0.906304 )
v = 1.904c / 1.906304
v = 0.9988c
Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c
Answer:
When you look at a simple koi pond you can find Koi (the secondary consumer) that feeds off of the zooplankton (first consumer), they eat the phytoplankton (producers). All in a simple food chain
Explanation:
Basically, Koi eat the little animal plankton (zooplankton) that then eats the plant plankton (phytoplankton) that can only end when a part of that habitat is removed. If you got rid of the plant plankton then the whole chain would collapse and most likely die.
Answer:
C.) 1
Explanation:
1 is the closest to 1.04
Also you never want to go above your prescription amount
