A cylindrical container with a cross sectional area of 65.2 cm^2 holds a fluid of density 806 kg/m^3. At the bottom of the conta
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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°
The electric potential energy of the charge is equal to the potential at the location of the charge, V, times the charge, q:
The potential is given by the magnitude of the electric field, E, times the distance, d:
So we have
(1)
However, the electric field is equal to the electrical force F divided by the charge q:
Therefore (1) becomes
And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
The potential is given by the magnitude of the electric field, E, times the distance, d:
So we have
However, the electric field is equal to the electrical force F divided by the charge q:
Therefore (1) becomes
And if we use the data of the problem, we can calculate the electrical potential energy of the charge: