Answer:
The frictional force 6.446 N
The acceleration of the block a = 6.04
Explanation:
Mass of the block = 3.9 kg
°
= 0.22
(a). The frictional force is given by
3.9 × 9.81 ×
29.3 N
Therefore the frictional force
0.22 × 29.3
6.446 N
(b). Block acceleration is given by
F = 30 N
= 6.446 N
= 30 - 6.446
= 23.554 N
The net force acting on the block is given by
23.554 = 3.9 × a
a = 6.04
This is the acceleration of the block.
The change in potential along a path from C to D due to a small charged sphere is 900 V.
Given:
Charge, Q = 3 nC = 3 × 10⁻⁹ C
Distance between the sphere and point C, r₁ = 0.02 m
Distance between the sphere and point D, r₂ = 0.06 m
Calculation:
We know that the electric potential is given as:
V = k Q/r - (1)
where, V is the electric potential
k is Coulomb's force constant
Qis the charge on the sphere
r is the separation distance
The electric potential at point C due to charged sphere can be given as:
V₁ = k Q/r₁
= (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.02 m)]
= 1350 V
The electric potential at point D due to charged sphere can be given as:
V₂ = k Q/r₂
= (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.06 m)]
= 450 V
Now, the change in potential along the path from C to D can be calculated as:
ΔV = V₂ - V₁
= 450 V - 1350 V
= -900 V
The negative sign indicates that the work is done against the electric field in moving the charge from C to D.
Therefore, the change in potential along a path from C to D is 900 V against the direction of the electric field.
Learn more about the electric potential here:
<u>brainly.com/question/12645463</u>
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