You have a 2.00 kg block of lead. Lead melts at 327.5°C. Cpb = 130.0 J/kg*C and △Hj for lead is 2.04x10^4 J/kg . Say you start a
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Answer:
Explanation:
let this sign be ∅ (titha in degrees)
let coefficient of friction of 0.200 be u
and i be current i= 1.41A
mass m=1.14g/cm m=0.0114g/m
To find the magnetic field from the north will be:
tan ∅= u
∅=
∅=11.3°
The direction:
i L*B cos(∅)= umg
divide both by iLcos(∅) to find B
B= where m=
B=
B=
Answer:
Explanation:
A) In order to solve the table it is necessary to consult tables A11-E and A10E for refrigerant R134-a
In this way we obtain that:
In this way,
In this way the entropy change is,
B) Whenever entropy yields a positive result, the process can be carried out adiabatically.