Answer:
Explanation:
According to the free-body diagram of the system, we have:
So, we can solve for T from (1):
Replacing (3) in (2):
The electric force () is given by the Coulomb's law. Recall that the charge q is the same in both spheres:
According to pythagoras theorem, the distance of separation (r) of the spheres are given by:
Finally, we replace (5) in (4) and solving for q:
Answer:
95.9°
Explanation:
The diagram illustrating the action of the two forces on the object is given in the attached photo.
Using sine rule a/SineA = b/SineB, we can obtain the value of B° as shown in the attached photo as follow:
a/SineA = b/SineB,
83/Sine52 = 56/SineB
Cross multiply to express in linear form
83 x SineB = 56 x Sine52
Divide both side by 83
SineB = (56 x Sine52)/83
SineB = 0.5317
B = Sine^-1(0.5317)
B = 32.1°
Now, we can obtain the angle θ, between the two forces as shown in the attached photo as follow:
52° + B° + θ = 180° ( sum of angles in a triangle)
52° + 32.1° + θ = 180°
Collect like terms
θ = 180° - 52° - 32.1°
θ = 95.9°
Therefore, the angle between the two forces is 95.9°
Answer:
Yes, if the system has friction, the final result is affected by the loss of energy.
Explanation:
The result that you are showing is the conservation of mechanical energy between two points in the upper one, the energy is only potential and the lower one is only kinetic.
In the case of some type of friction, the change in energy between the same points is equal to the work of the friction forces
= ΔEm
=
-Em₀
As we can see now there is another quantity and for which the final energy is lower and therefore the final speed would be less than what you found in the case without friction.
=
+ Em₀
Remember that the work of the rubbing force is negative, let's write the work of the rubbing force explicitly, to make it clearer
½ m v² = -fr d + mgh
v = √(-fr d 2/m + 2 gh)
v = √ (2gh - 2fr d/m)
Now it is clear that there is a decrease in the final body speed.
Consequently, if the system has friction, the final result is affected by the loss of energy.