The value of the final speed depends on the mass of the ore.
Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.
Initially, there is only the cart, so the momentum is
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is
. The momentum p is conserved, so it is still 12960 kg m/s. Therefore, we have
and so the final speed is
-6 uh 89 0 -+= 62 uwuwhhiwhiuhwiuhwuihhishishoknsk
Answer:
0.75
Explanation:
Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.
Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).
So, μ = F/N
= 300 N/400 N
= 3/4
= 0.75
So, the coefficient of static friction μ = 0.75
(a) 
According to Newton's second law, the force experienced by each balloon is given by:
F = ma
where
m = 0.021 kg is the mass
a = 1.1 m/s^2 is the acceleration
Substituting, we found:
The electrostatic force between the two balloons can be also written as
where
k is the Coulomb's constant
Q is the charge on each balloon
r = 16 m is their separation
Since we know the value of F, we can find Q, the magnitude of the charge on each balloon:
(b) 
electrons
The magnitude of the charge of one electron is
While the magnitude of the charge on one balloon is
This charge can be written as
where N is the number of electrons that are responsible for this charge. Solving for N, we find:
Answer: I think, the number of oscillations in a given period of time.
Explanation: Well I guess because in a period time is known as the rate of occurrence of the wave. Hope this helps!
