Answer:
13.98 nC
Explanation:
Capacitance depends upon the area of the plates and their distance of separation.
Radius = r = 0.071 m
separation = d = 0.00126 m
here κ = 1 and ε₀ = 8.85 ₓ 10⁻¹² SI units , for free space.
Area = A = π r² = 0.0158 m²
C = [( 8.85 ₓ 10⁻¹² ) ( 0.0158) ]÷ (0.00126) = 1.11 x 10⁻¹⁰ F
Charge = Q = C V = ( 1.11 x 10⁻¹⁰ F )(126) = 13.98 nC
= 14 nC ( rounded to two significant digits)
Answer:
The amplitude of the oscillation is 2.82 cm
Explanation:
Given;
mass of attached block, m = 4.1 kg
energy of the stretched spring, E = 3.8 J
period of oscillation, T = 0.13 s
First, determine the spring constant, k;
where;
T is the period oscillation
m is mass of the spring
k is the spring constant
Now, determine the amplitude of oscillation, A;
where;
E is the energy of the spring
k is the spring constant
A is the amplitude of the oscillation
Therefore, the amplitude of the oscillation is 2.82 cm
The given question is incomplete. The complete question is as follows.
A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.
Explanation:
The given data is as follows.
= 20 N,
= 25 N, a = -0.9
W = 83 N
m =
= 8.46
Now, we will balance the forces along the y-component as follows.
N = W +
= 83 + 25 = 108 N
Now, balancing the forces along the x component as follows.
= ma
= 7.614 N
Also, we know that relation between force and coefficient of friction is as follows.
=
= 0.0705
Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.