A 300-kg bear grasping a vertical tree slides down at constant velocity. the friction force between the tree and the bear is:___
1 answer:
3000 N of friction exists between the bear and the tree.
The bear goes down at a constant speed, thus there is no acceleration. The forces add up to zero. As a result, upward friction and downward weight result in zero
i.e.,
F-mg=0
F = mg
F = (300 kg)/10 (gravitational acceleration)
3000 N frictional force equals F.
How does friction force work?
- Friction is the force that prevents two solid objects from rolling or sliding over one another. Even though frictional forces, such as the traction required to walk without slipping, may be advantageous, they can also be a significant hindrance to motion.
Between solid surfaces, there are three basic types of friction:
- Rolling
- Sliding.
- Static. They are graded from strongest to weakest. Fluid friction happens when liquids or gasses are mixed together.
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Here is the missing part of the question
To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C
Answer:
Explanation:
The image attached below shows the process on T - v diagram
<u>At State 1:</u>
The first step is to find the vapor pressure
= 0.5 × 232 kPa
= 116 kPa
The initial specific volume of the vapor is:
<u>At State 1:</u>
The next step is to determine the mass of water vapor pressure.
= 1.578 kg
Using the ideal gas equation to estimate the mass of the dry air
For the specific volume , we get the identical value of saturation temperature
Thus, at , condensation needs to begin.
However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.
Heat can now be determined by using the formula
Q = ΔU + W
Recall that: For a rigid tank, W = 0
Q = ΔU + 0
Q = ΔU
Q = U₂ - U₁
Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.
<u>At State 1;</u>
The internal energy is calculated as:
At = 125° C, we obtain the specific internal energy of air
SO;
At = 125° C, we obtain the specific internal energy of water vapor
= 6.216 × 284.714 + 1.578 × 2534.5
= 5768.716 kJ
<u>At State 2:</u>
The internal energy is calculated as:
At temperature 110° C, we obtain the specific internal energy of air
SO;
At temperature 110° C, we obtain the specific internal energy of water vapor
= 6.216 × 273.862 + 1.578 × 2517.9
= 5675.57 kJ
Finally, the heat transfer during the process is
Q = U₂ - U₁
Q = (5675.57 - 5768.716 ) kJ
Q = -93.146 kJ
with the negative sign, this indicates that heat is lost from the system.
