Answer:
C Quincy
Explanation:
Both Frieda and Vladimir are too young to be making their own decisions. Lucinda has limited freedom due to having two children, while Quincy is just now becoming an adult and has his whole life still ahead of him. Therefore, Quincy has the most freedom to make their own lifestyle decisions.
I hope this helps!
Answer:
1. 3 m
2. 27 s
Explanation:
1. "A car traveling at +33 m/s sees a red light and has to stop. If the driver can accelerate at -5.5 m/s², how far does it travel?"
Given:
v₀ = 33 m/s
v = 0 m/s
a = -5.5 m/s²
Unknown: Δx
To determine the equation you need, look for which variable you don't have and aren't solving for. In this case, we aren't given time and aren't solving for time. So look for an equation that doesn't have t in it.
Equation: v² = v₀² + 2aΔx
Substitute and solve:
(0 m/s)² = (33 m/s)² + 2(-5.5 m/s²) Δx
Δx = 3 m
2. "A plane starting from rest at one end of a runway accelerates at 4.8 m/s² for 1800 m. How long did it take to accelerate?"
Given:
v₀ = 0 m/s
a = 4.8 m/s²
Δx = 1800 m
Unknown: t
Equation: Δx = v₀ t + ½ a t²
Substitute and solve:
1800 m = (0 m/s) t + ½ (4.8 m/s²) t²
t ≈ 27 s
Copper, because it has the lowest specific electrical resistance.
specific electrical resistance aka volume resistivity is a fundamental property of a material that quantifies how strongly that material opposes the flow of electric current. A low resistivity indicates a material that readily allows the flow of electric current.
Answer:
- Decreasing the resistance
- Using a shorter length
- Using a smaller area wire
Explanation:
Formula for conductance in wires is;
G = 1/R
Where;
G is conductance
R is resistance
This means that increasing the resistance leads to a larger denominator and thus a smaller conductance but to decrease the denominator means larger conductance.
Thus, to increase the conductance, we have to decrease the resistance.
Resistance here has a formula of;
R = ρL/A
Where;
ρ is resistivity
L is length of wire
A is area
Thus, to decrease the resistance, we will have to use a shorter length and smaller area of wire.
