Refer to the diagram shown below.
The given data is
mass, kg Coordinates. m
------------- -----------------
2 (0, 0)
2 (2, 0)
4 (2, 1)
Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.
Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m
8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m
Answer: (1.5, 0.5) m
16/9 m/s^2
negative 4/3 m/s^2
14 m/s
the last one is too detailed to do in my head while on the bus; sorry
a) Sketches of all possible pv-diagrams for the cycle are attached below
b) The work W
for the process Ca is : 2462.8 J
<u>Given data :</u>
Amount of heat flowing out = 800 J
Ta = 200 K
Tb = 300 K
R = 800
<u>B) Determine the </u><u>work W </u><u>for the process</u><u> Ca</u><u> </u>
Wₐs = -pdv
= - [ pVb - pVa ] ---- ( 1 )
note : pVb = nRTb , pVa = nRTa
Equation ( 1 ) becomes
= -nR [ Tb - Ta ]
= - 2(8.314 ) [ 300 - 200 ]
= - 1662.87
given that W
= 0 which is isochonic
dv = 0 ( cyclic process ) = d∅ - dw
∴ 0 = 800 - ( Wₐs + W )
Therefore : W = 800 + 1662.8 = 2462.8 J
Hence we can conclude that the work W for the process Ca = 2462.8 J
Learn more about Pv diagrams : brainly.com/question/25401637
Answer:
200
Explanation:
has to be long this us just random words anseer is b. 100
I believe this question ask for the energy dissipated by
friction.
The overall energy equation for this is:
F = PE – KE
where F is friction loss, PE is potential energy = m g h,
KE is kinetic energy = 0.5 m v^2
<span>F = 66 kg * 9.8 m/s^2 * 170 m – 0.5 * 66 kg * (11 m/s)^2</span>
<span>F = 105,963 J ~ 106,000 J </span>
