Answer: 10m/s^2
Explanation:
Given the following :
Initial Velocity(U) = 10m/s
Angle of inclination = 30°
In other to calculate the magnitude of the vector,
We need to resolve into verti AL and horizontal component.
Vertical component of a vector(Uy) = U × Sinθ
Horizontal component(Ux) = U × Cosθ
Uy = 10 × sin30 = 10 × 0.5 = 5m/s
Ux = 10 × Cos30 = 10 × 0.866 = 8.66m/s
Therefore the magnitude equals :
U = √(Uy)^2 + (Ux)^2
U = √5^2 + 8.66^2
U = √25 + 74.9956
U = √99.9956
U = 9.9997799
U = 10m/s^2
Answer:
200 J
Explanation:
Work = Force × Distance
Work = (20 N)(10 m) = 200 N·m = 200 J
200 joules of work are done by a force of 20 N over a distance of 10 m.
Answer:
the maximum current is 500 A
Explanation:
Given the data in the question;
the B field magnitude on the surface of the wire is;
B = μ₀i / 2πr
we are to determine the maximum current so we rearrange to find i
B2πr = μ₀i
i = B2πr / μ₀
given that;
diameter d = 2 mm = 0.002 m
radius = 0.002 / 2 = 0.001 m
B = 0.100 T
we know that permeability; μ₀ = 4π × 10⁻⁷ Tm/A
so we substitute
i = (0.100)(2π×0.001 ) / 4π × 10⁻⁷
i = 500 A
Therefore, the maximum current is 500 A
