A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
nuclear fusion- produces huge amounts of energy.
The displacement is the shortest distance between two points, which is 546.41. The displacement for both is 546.41 meters
Average velocity of X = (200 + 200 + 200) / 30
Average velocity of X = 20 m/s
Average velocity of Y = 546.41 / 30 = 18.2 m/s
Answer:
d. decreases
Explanation:
The law of conservation of momentum tells us that the sum of momenta before the collision is equal to the sum of momenta after the collision. The bag has no momentum as it falls onto the boat because its velocity is zero in the horizontal direction. But after it hits the boat, it's momentum increases while the momentum of the system remains the same. That means a component of the system must decrease somewhere else. And that component is the velocity, not the mass, of the boat.
