Answer:
80 m/s
Explanation:
Given:
a = -5 m/s²
v = 0 m/s
Δx = 640 m
Find: v₀
v² = v₀² + 2a(x − x₀)
(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)
v₀ = 80 m/s
Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
Answer:
Explanation:
= Length of wire = 65 m
= Initial current = 1.8 A
= Final current = 2.9 A
We know
and
so
The length of the wire remaining on the spool is .
Answer:
sorry i don't no
i promise i will help you later
now i am also in trouble now
nobody helps me
Gravity is the only one, since there's no air resistance.
