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
Catalysts are used in the oil industry to increase octane
ratings in gasoline. Octane ratings tell how much fuel has to be compressed to
promote better fuel ignition in some
cars. The catalytic effect that
would be most beneficial for the oil industry is that more products can be
formed from a given amount of material.
"The are many different seismic waves, but all of basically of four types:
Compressional or P (for primary)
or S (for secondary)
Love
Rayleigh
An earthquake radiates P and S waves in all directions and the interaction of the P and S waves with Earth's surface and shallow structure produces surface waves."
Evidence for the particle nature of light are not: 1. refraction, 2. many colors of light, 3. diffraction. These are all phenomenon that support wave theory of light. Evidence for particle nature of light is photoelectric effect. Because it was discovered that you need discrete energies of light to eject electrons from a metal surface and not continuous as the wave theory of light suggests.
Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
