Answer:
The force has been reduced by 8018 N
Explanation:
The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:
where:
F is the force exerted on the car
is the duration of the collision
m = 1400 kg is the mass of the car
is the change in velocity of the car
We can re-write the equation as
In the 1st collision, the time is 1.5 seconds, so the force is
In the 2nd collision, the time is increased to 2.2 seconds, so the force is
Therefore, the force has been reduced by:
I believe it is -1.11 m/s^2. I will let you know if its correct
Answer:
Explanation:Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of, ... This article is about the physical phenomenon. ... If the diameter of the tube is sufficiently small, then the combination of surface tension (which is caused by cohesion ... They derived the Young–Laplace equation of capillary action.
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
