(6-16)/4.0=-2.5 m/s²
Acceleration of the car is -2.5 m/s²
Since the system itself is giving off heat, this is a
reduction in the internal energy.
heat = - 25,000 J
Since work is being done on the system, therefore it is
an additional energy to the system. Work is given as:
work = - P dV
work = - 1.50 atm (6 L – 12 L)
work = 9 L atm
Since it is given that 1 L atm is equivalent to 101.3 J,
therefore the total energy added is:
energy due to work = 9 L atm (101.3 J / 1 L atm)
energy due to work = 911.7 J
Therefore the total change in internal energy is the sum
of heat and energy due to work:
Change in internal energy = - 25,000 J + 911.7 J
Change in internal energy = - 24,088.3 J
<span>Therefore, approximately 24.1 kJ of energy is lost by the
system in the total process.</span>
<span>
</span>
<span>Answer:</span>
<span>-24.1 kJ</span>
Answer:
I assume by "which" of these you're looking for an example. Water freezing into ice or water, or evaporation, the process of turning from liquid into vapor, would not be chemical changes.
Explanation:
These are physical changes because they do not form a new substance, a chemical change requires a change in the chemical makeup of the substance.
I can't seem to figure out the angle between T1 and T2. So suppose, it is 10º; then T2 makes an angle of 35º w/r/t horizontal, and T1 makes an angle of 45º.
Sum the moments about the base of the crane; Σ M = 0. 0 = T2*cos35*L*cos40 + T1*cos45*L*cos40 - T2*sin35*L*sin40 - T1*sin45*L*sin40 - W*(L/2)*sin40 - T1*L*sin40 → length L cancels where W = 18 kN
0 = 0.259*T2 - 43kN T2 = 166 kN
