Potential energy is the energy stored in matter because of
its position or because of the arrangement of parts.
Answer:
the car will slow down due to the change in speed
Answer:
24,000 m
Explanation:
First find the rocket's final position and velocity during the first phase in the y direction.
Given:
v₀ = 75 sin 53° m/s
t = 25 s
a = 25 sin 53° m/s²
Find: Δy and v
Δy = v₀ t + ½ at²
Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²
Δy = 7736.8 m
v = at + v₀
v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)
v = 559.0 m/s
Next, find the final position of the rocket during the second phase (as a projectile).
Given:
v₀ = 559.0 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 15945.5 m
The total displacement is:
7736.8 m + 15945.5 m
23682.2 m
Rounded to two significant figures, the maximum altitude reached is 24,000 m.
1- first law
2- third law
3- first law
4- second law
5- third law
6- second law
Answer:
12900 W
24200 W
Explanation:
Given:
v₀ = 0 m/s
v = 1.3 m/s
t = 2.0 s
Find: a and Δx
v = at + v₀
(1.3 m/s) = a (2.0 s) + (0 m/s)
a = 0.65 m/s²
Δx = ½ (v + v₀) t
Δx = ½ (1.3 m/s + 0 m/s) (2.0 s)
Δx = 1.3 m
While accelerating:
Newton's second law:
∑F = ma
F − mg = ma
F = m (g + a)
F = (1500 kg + 400 kg) (9.8 m/s² + 0.65 m/s²)
F = 19855 N
Power = work / time
P = W / t
P = Fd / t
P = (19855 N) (1.3 m) / (2.0 s)
P ≈ 12900 W
At constant speed:
Newton's second law:
∑F = ma
F − mg = 0
F = mg
F = (1500 kg + 400 kg) (9.8 m/s²)
F = 18620 N
Power = work / time
P = W / t
P = Fd / t
P = Fv
P = (18620 N) (1.3 m/s)
P ≈ 24200 W