Answer:
The electric field produced by this disk along the x axis at point (P = 1.01 m, 0.00 m) is 996.54 N/C
Explanation:
The electric field produced by this disk along the x axis at point (P = 1.01 m, 0.00 m), will be evaluated as follows:
Since x > 0
![E_{x} = 2\pi \sigma k [1-\frac{x}{\sqrt{x^2 +R^2}}]](https://tex.z-dn.net/?f=E_%7Bx%7D%20%3D%202%5Cpi%20%5Csigma%20k%20%5B1-%5Cfrac%7Bx%7D%7B%5Csqrt%7Bx%5E2%20%2BR%5E2%7D%7D%5D)
σ is surface charge density = 5.88 × 10⁻⁶ C / m²
R is the radius = 7.52cm = 0.0752m
position x = 1.01m
k is coulomb force constant = 8.99 × 10⁹ Nm² / C²
![E_{x} = 2\pi ( 5.88X10^{-6})(8.99 X10^9)[1-\frac{1.01}{\sqrt{1.01^2 +0.0752^2}}]](https://tex.z-dn.net/?f=E_%7Bx%7D%20%3D%202%5Cpi%20%28%205.88X10%5E%7B-6%7D%29%288.99%20X10%5E9%29%5B1-%5Cfrac%7B1.01%7D%7B%5Csqrt%7B1.01%5E2%20%2B0.0752%5E2%7D%7D%5D)
= 996.54 N/C
Therefore, The electric field produced by this disk along the x axis at point (P = 1.01 m, 0.00 m) is 996.54 N/C
An atom would be your answer, so B!
Answer:
F = 768 N
Explanation:
It is given that,
Speed of the elevator, v = 3.2 m/s
Grain drops into the car at the rate of 240 kg/min, 
We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :
Since, the speed is constant,
F = 768 N
So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.
The net force applied to the object equals the mass of the object multiplied by the amount of its acceleration." The net force acting on the soccer ball is equal to the mass of the soccer ball multiplied by its change in velocity each second (its acceleration).
Answer:
Explanation:
<u>Friction Force</u>
When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.
There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.
Please find the free body diagrams in the figure provided below.
The equilibrium condition for the mass 1 is
The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa
![\displaystyle F_a=F_{r1}+F_{r2}.....[1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_a%3DF_%7Br1%7D%2BF_%7Br2%7D.....%5B1%5D)
The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is
The friction forces are computed by
Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.
Replacing in [1]
Simplifying
Plugging in the values
![\displaystyle F_{a}=0.25(9.8)[400+2(100)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_%7Ba%7D%3D0.25%289.8%29%5B400%2B2%28100%29%5D)
