The net force on the object as described is; 58.84N
Two forces acting on the object are;
- The <em>applied force and the frictional force.</em>
In essence; the frictional force can be evaluated as;
- Frictional force; = coefficient × Weight of object.
- Frictional force = 0.21 × 20 × 9.8.
- Frictional force = 41.16N
- The Net force = Applied force - frictional force
Net Force = 58.84 N.
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Answer:
Using the log combination rules to reduce the famous Sakur-Tetrode equation, The change in entropy is given as:
∆S = NK*ln(V"V$/V").
Where V"V$ is final Volume (Vf) after constraint's removal,
V" is Initial Volume (Vi) before constraint's removal.
Temperature (T) is constant, Internal Energy, U is constant, N and K have their usual notations
Explanation:
Given in the question, the container is an adiabatic container.
For an adiabatic contain, it does not permit heat to the environment due to its stiff walls. This implies that the Internal Energy, U is kept constant(Q = U). The temperature is also constant (Isothermal). Thus, the famous Sakur-Tetrode equation will reduce to ∆S = NK* In(Vf/Vi).
Vf is the volume after the constraint is removed(Vf = V"V$). Vi is the volume occupied before the constraint is removed (Vi = V")
Answer:
Distance, d = 3242.19 meters
Explanation:
It is given that,
Speed of the dog, v = 19.85 mi/hr
Since, 1 mph = 0.44704 m/s
v = 19.85 mi/hr = 8.873 m/s
Time taken by the dog, t = 6.09 min = 365.4 sec
Let the distance covered by the dog during this time period is d. It can be calculated by the speed of the dog multiplied by the time taken as :
d = 3242.19 meters
So, the the dog travel during this time period is 3242.19 meters. Hence, this is the required solution.
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Answer:
The Rock
Explanation:
The paper is very thin and light so it has more air reisistance which will cause it to fall slower than the rock that has a larger density.
