Assuming that the can is motionless, we can then assume that the vertical component of T = mg and that Fe = the horizontal component of T.
<span> Since T itself is larger than it's vertical or horizontal components separately, then T is greater than all the forces.</span>
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Answer:
1. True WA > WB > WC
Explanation:
In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.
A) Work is the product of force by distance and the cosine of the angle between them
WA = W h cos 0
WA = mg h
B) On a ramp without rubbing
Sin30 = h / L
L = h / sin 30
WB = F d cos θ
WB = F L cos 30
WB = mf (h / sin30) cos 30
WB = mg h ctan 30
C) Ramp with rubbing
W sin 30 - fr = ma
N- Wcos30 = 0
W sin 30 - μ W cos 30 = ma
F = W (sin30 - μ cos30)
WC = mg (sin30 - μ cos30) h / sin30
Wc = mg (1 - μ ctan30) h
When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum
Let's review the claims
1. True The work of gravity is the greatest and the work where there is friction is the least
2 False. The job where there is friction is the least
3 False work with rubbing is the least
4 False work with rubbing is the least
Answer:
The potential difference between the plates increases
Explanation:
As we know that the capacitance of the capacitor is given by:
(1)
where
q = charge
C = capacitance
V = Voltage or Potential Difference
Also, the capacitance of a parallel plate capacitor is given as:
(2)
where
A = Area of the plates
D = Separation distance between the plates
Now, from eqn (1) and (2):
Now, from the above eqn we can say that:
Potential difference depends directly on the separation distance between the plates of the capacitor and is inversely dependent on the area of the plates of the capacitor.
Therefore, after disconnecting, if the separation between the plates is increased the potential difference across it also increases.
Radiation is a method of heat transfer through electromagnetic waves. This transfer does not require a medium in order for heat to be transferred.
The solution is A.
