yes the car is chaning direction throught the track, so it is accelerting.
The answer is D) because the speeds are different for each phase of matter!
Remember that since energy is conserved, the kinetic energy it takes to jump to a height of 1.2 m is just the same as the potential energy difference between ground level and the 1.2 m height.
Potential energy = (mass)(gravity)(height) = (7.2 kg)(9.81 m/s^2)(1.2 m) = 84.76 J
Therefore, the kinetic energy required for the dog to jump to a 1.2-m height is 84.76 Joules.
C. 0.37V. A capacitor of 650x10⁻⁴F that stores 24x10⁻³C has a potential difference of 0.37V between its plates.
The key to solve this problem is using the capacitance equation C = Q/Vᵃᵇ, where C is the capacitance, Q the charge stored in the plates, and Vᵃᵇ the potential difference between the plates.
A 650x10⁻⁴F capacitor stores 24x10⁻³C, clear Vᵃᵇ for the equation:
C = Q/Vᵃᵇ -----------> Vᵃᵇ = Q/C
Solving
Vᵃᵇ = 24x10⁻³C/650x10⁻⁴F = 0.37V
Answer:
66.375 x 10⁻⁶ C/m
Explanation:
Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as
;
∅ = Q / ε₀ -----------------(i)
Where;
∅ = 7.5 x 10⁵ Nm²/C
ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²
Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;
7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)
Solve for Q;
Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²
Q = 66.375 x 10⁻⁷ C
Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e
L = Q / l ----------------------(ii)
Where;
Q = 66.375 x 10⁻⁷ C
l = length of the rod = 10.0cm = 0.1m
Substitute these values into equation (ii) as follows;
L = 66.375 x 10⁻⁷C / 0.1m
L = 66.375 x 10⁻⁶ C/m
Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.
