With countercurrent flow, diffusion happened in all regions of the filter. Explain why
1 answer:
You might be interested in
The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
The magnitude of momentum of the airplane is .
Explanation:
Given that,
Mass of the airplane, m = 3400 kg
Speed of the airplane, v = 450 miles per hour
Since, 1 mile per hour = 0.44704 m/s
v = 201.16 m/s
We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,
or
So, the magnitude of momentum of the airplane is . Hence, this is the required solution.