Answer:
7.6 g
Explanation:
"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.
The heat gained by the copper, water, and ice = the heat lost by the steam
Heat gained by the copper:
q = mCΔT
q = (120 g) (0.40 J/g/K) (40°C − 0°C)
q = 1920 J
Heat gained by the water:
q = mCΔT
q = (70 g) (4.2 J/g/K) (40°C − 0°C)
q = 11760 J
Heat gained by the ice:
q = mL + mCΔT
q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)
q = 4880 J
Heat lost by the steam:
q = mL + mCΔT
q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)
q = 2452 J/g m
Plugging the values into the equation:
1920 J + 11760 J + 4880 J = 2452 J/g m
18560 J = 2452 J/g m
m = 7.6 g
Answer:
It’s 7 hours
Explanation:
You have to use the formula your teacher has given to you plug in the numbers then solve be sure to use a calculator made for physics it helps a lot :)
Where is the diagram? What is the question?
At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
<h3>What is the energy of the roller coaster at point E?</h3>
The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.
Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.
At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,
Learn more about potential and kinetic energy at: brainly.com/question/18963960
#SPJ1
Solution :
Given data :
Mass of the merry-go-round, m= 1640 kg
Radius of the merry-go-round, r = 7.50 m
Angular speed, 
rev/sec
rad/sec
= 5.89 rad/sec
Therefore, force required,
= 427126.9 N
Thus, the net work done for the acceleration is given by :
W = F x r
= 427126.9 x 7.5
= 3,203,451.75 J
