Answer:
a) a = 3.06 10¹⁵ m / s
, b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m
- m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N
Constant acceleration of plane = 3m/s²
a) Speed of the plane after 4s
Acceleration = speed/time
3m/s² = speed/4s
S = 12m/s
The speed of the plane after 4s is 12m/s.
b) Flight point will be termed as the point the plane got initial speed, u, 20m/s
Find speed after 8s, v
a = 3m/s²
from,
a = <u>v</u><u> </u><u>-</u><u> </u><u>u</u>
t
3 = <u>v</u><u> </u><u>-</u><u> </u><u>2</u><u>0</u>
8
24 = v - 20
v = 44m/s
After 8s the plane would've 44m/s speed.
Answer:
hot I checked already in my phone