Hopes this helps:
Answer: Aluminum has 61 percent of the conductivity of copper, but has only 30 percent of the weight of copper. That means that a bare wire of aluminum weights half as much as a bare wire of copper that has the same electrical resistance. Aluminum is generally more inexpensive when compared to copper conductors.
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
GPE= height x mass x gravitational field strength
5 x 10 x 9,8=490J
Both are constants used in the definition of Forces (gravitational and electric,respectively)
Since those constants are proportional to the magnitude of the forces:
Having a small gravitational constant explains why there is no apparent force of attraction with objects of considerable low mass (they would need to have great value of mass for the equation to give an apreciable force)
Electrical interactions are usually strong, and thus require an appropiate constant to depict the phenomenon. We deal in this case with charges really small, but the forces are in different order of magnitude.
Answer:
In a series circuit, the same amount of current flows through all the components placed in it. On the other hand, in parallel circuits, the components are placed in parallel with each other due to which the circuit splits the current flow.