Answer:
The answer is below
Explanation:
a) The vertical displacement = Δy = 21.5 m - 1.5 m = 20 m
The horizontal displacement = Δx = 69 m wide
Using the formula:

Also:

b)The car is moving at a constant speed in the horizontal direction, hence the initial velocity = final velocity

The answer is speed: 4.7 km/h, velocity: 3.3 km/h.
Distances and time are given:
d1 = 4 km
d2 = 3 km
d3 = 5 km
t = 1.5 h
The speed can be expressed as a distance (d) divided by time (t). The average speed (s) is total distance travelled divided by time:
s = (d1 + d2)/t = (4+3)/1.5 = 7/1.5 = 4.7 km/h
The average velocity (v) is total displacement (d₁) from the starting point divided by time. Since Mary's starting point was home, and she walked to the supermarket, which is 5.0 kilometers from her own home, her displacement is 5 km:
v = d₁/t = 5/1.5 = 3.3 km/h
Answer:

Explanation:
From the law of conservation of energy
Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr



The required distance from A to B is 
Answer:
115 kPa
Explanation:
Use Bernoulli equation:
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
Assuming no elevation change, h₁ = h₂.
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²
Plugging in values:
(582,000 Pa) + ½ (1000 kg/m³) (1.28 m/s)² = P + ½ (1000 kg/m³) (30.6 m/s)²
P = 115,000 Pa
P = 115 kPa
Hey
<span>.0081/300 = .000027 = 2.7 x 10^-5
</span>
Hoped I Helped