Internal energy, U, is equal to the work done or by the system, plus the heat of the system:
<span>ΔU=q+w
</span>in the question they tell you the work done by the system, and the internal energy:
8185 J= -346 J + q work is negative because it was done BY the system.
substitute in: <span>q=m∗Cp∗ΔT</span> and solve for <span>Cp</span><span>.
</span>
-------------------------------------
remember that <span>ΔT=<span>Tf</span>−<span>Ti
</span></span>
so the equation, really, is: <span>q=m∗Cp∗(<span>Tf</span>−<span>Ti</span>)</span><span>
------------------------------------------
</span>
<span>185J=−346J+[m∗Cp∗(<span>Tf</span>−<span>Ti</span>)]
</span>plug in the rest of your values and solve for <span><span>Cp</span></span>
Answer:
The change in potential energy of the mass as it goes up the incline is 0.343 joules.
Explanation:
We must remember in this case that change in the potential energy is entirely represented by the change in the gravitational potential energy. From Work-Energy Theorem and definition of work we get that:
Where:
- Gravitational potential energy, measured in Joules.
- Mass, measured in kilograms.
- Gravitational acceleration, measured in meters per square second.
- Change in vertical height, measured in meters.
This work is the energy needed to counteract effects of gravity at given vertical displacement.
If we know that 
, 
and 
, the change in the potential energy of the mass as it goes up the incline is:
The change in potential energy of the mass as it goes up the incline is 0.343 joules.
Answer:
4.03 m/s
Explanation:
Initial momentum = final momentum
(282 kg) (3.50 m/s) + (155 kg) (-1.38 m/s) = (282 kg) (1.10 m/s) + (115 kg) v
v = 4.03 m/s
Answer:
Focal length f=−20 cm
Object distance u=−50 cm
Using
v
1
−
u
1
=
f
1
Or
v
1
−
(−50)
1
=
−20
1
We get v=−14.3 cm
Since, v is negative. So image formed is virtual.
Hence, virtual image is formed at a distance of 14.3 cm in front of the lens.
A)The speed of the rider is 2.6m/s
B)The rider's radial acceleration is 
C)The radial acceleration when the time for one rotation is halved is 
Explanation:
Time for one rotation t=6s
speed is the rate of change of distance with time.
to calculate the distance we have to find the circumference of the ride
A)
B)
C)If the time of rotation is halved, speed/velocity as well as the radial acceleration changes.
new time t'=3s
new velocity v'=circumference/t'=15.4/3=5.13 m/s
