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
Answer:
Energy is force times distance. For your problem, no matter how long you push, the wall still goes nowhere, so there is no obvious energy transfer. so in conclusion, you actually didn't do anything :(
Explanation:
Yes. It means that the acceleration increases at a constant rate, for example 3 mph every second.
Answer:
2m/s/s
Explanation:
The formula goes- F=MA
F-Force M-Mass & A-Acceleration
We need to rearrange this formula to find the acceleration-
A=F/M
All we need to do now is substitute the values in
A=2000N/1000kg
A=2m/s^2
In the given option the last option (2m/s/s) would be the ans, as it's the same as 2m/s^2
So ya, I guess that's all