Answer:
Explanation:
Given
mass of First Block 
Temperature 
mass of second block 
Temperature 
Heat capacity of aluminium c=899 J/kg-K
Final Temperature acquired by both blocks at steady state
Heat loss first block =Heat gain by second block




Answer: Got It!
<em>Explanation: </em>let s = speed at launch
v = 0 at top = s sin 63 - g t
so at top
t = s sin 63/g = .0909 s
h = 13.6 = s sin 63 t - 4.9 t^2
13.6 = .081s^2 - .0405 s^2
s^2 = 336
s = 18.3 m/s
0 0
Speed =dist./time
=73.4/5
=14.68 km/hr
Answer:
8672
Explanation:
multiply the length value by 100
Answer:
c.Beta (1 e-) is the answer.