Some examples of insulators are plastic, glass and rubber.
A. you're able to pull through facing out
B. you're able to park farther away from the building
C. you're able to reverse out of the parking spot
D. they are located in a parking garage
Option (A) is correct
Straight-in spaces can leave you a safer out if you're able to pull through facing out
Straight-in parking is a method to park a vehicle where the vehicle is guided and parked safely in between the guiding lines.
Each car is provided a separate slot for parking. It thus helps to prevent blockage of cars. Each car can move in and out freely without any congestion.
Straight-in parking is the most traditional approach of parking which saves time for drivers, allows for two-way traffic as well.
The driver can line up from multiple angles. But the safest way to leave out the Straight-in parking is to pull through facing out.
So option A is the most suitable option in the provided case.
Learn more about straight-in parking:
Answer: A. You're able to pull through facing out
Explanation: DriversED
B) decreases; increase; increases
C) increases; decrease; decreases
D) increases; increase; decreases
Calculate:
(a) the time to reach maximum height
(b) the maximum height above the base of the cliff reached by the
projectile
(c) thetotal time it is in the air
(d) the horizontal range of the projectile.
Answer:
a) 9.99 s
b) 538 m
c) 20.5 s
d) 1160 m
Explanation:
Given:
x₀ = 0 m
y₀ = 49.0 m
v₀ = 113 m/s
θ = 60.0°
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
a) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find t.
vᵧ = aᵧ t + v₀ᵧ
(0 m/s) = (-9.8 m/s²) t + (113 sin 60.0° m/s)
t ≈ 9.99 s
b) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find y.
vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)
(0 m/s)² = (113 sin 60° m/s)² + 2 (-9.8 m/s²) (y − 49.0 m)
y ≈ 538 m
c) When the projectile lands, y = 0 m. Find t.
y = y₀ + v₀ᵧ t + ½ aᵧ t²
(0 m) = (49.0 m) + (113 sin 60° m/s) t + ½ (-9.8 m/s²) t²
You'll need to solve using quadratic formula:
t ≈ -0.489, 20.5
Since negative time doesn't apply here, t ≈ 20.5 s.
d) When the projectile lands, y = 0 m. Find x. (Use answer from part c).
x = x₀ + v₀ₓ t + ½ aₓ t²
x = (0 m) + (113 cos 60° m/s) (20.5 s) + ½ (0 m/s²) (20.5 s)²
x ≈ 1160 m
2) paper burning
3) ice cream melting
4) a nail rusting
Answer: 3
Explanation: