Answer:
The Force between the two charges is an attractive force of 16,000N
Explanation:
Expression for the electric force between the two charges is given by
F = (k*q1*q2) / r^2
Here, k = constant = 9 x 10^9 N*m^2 / C^2
q1 = - 2.0x10^-4C
q2 = + 8.0x10^-4C
r = 0.30 m
Substitute the given values in the above expression -
One charge is + and the other is a -, therefore the net force is an attractive force (opposites atract)
The attraction force is:
F= 9.0x10^9 * 2.0x10^-4 *8.0x10^-4 N/ 0.30^2
F= 16,000N
In collision of the steel ball and the steel plate, the collision is an inelastic collision and there is loss in the kinetic energy.
<h3>What are collisions?</h3>
Collisions occur when two objects that are moving in the same directions or in different direction meet each other and collide.
There are two types of collisions:
- elastic collision - the kinetic energy is conserved
- inelastic collision - there is a loss in kinetic energy
In the collision of the steel ball and the steel plate, there is loss in the kinetic energy of the steel ball which is converted to sound energy.
In conclusion, the collision of the steel and steel plate is an inelastic collision.
Learn more about collisions at: brainly.com/question/7694106
#SPJ1
Answer:
Explanation:
From the given information:
We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then
The volume charge distribution relates to the radial direction at r = R
∴
![\rho (r) \ \alpha \ \delta (r -R)](https://tex.z-dn.net/?f=%5Crho%20%28r%29%20%5C%20%20%5Calpha%20%20%5C%20%20%5Cdelta%20%28r%20-R%29)
![\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)](https://tex.z-dn.net/?f=%5Crho%20%28r%29%20%3D%20k%20%5C%20%20%5Cdelta%20%28r%20-R%29%20%5C%20%5C%20%20at%20%5C%20%5C%20%20%28r%20%3D%20R%29)
![\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)](https://tex.z-dn.net/?f=%5Crho%20%28r%29%20%3D%200%5C%20%5C%20since%20%5C%20r%3C%20R%20%20%5C%20%5C%20or%20%20%5C%20%5C%20r%3ER----%20%281%29)
To find the constant k, we examine the total charge Q which is:
![Q = \int \rho (r) \ dV = \int \sigma \times dA](https://tex.z-dn.net/?f=Q%20%3D%20%5Cint%20%5Crho%20%28r%29%20%5C%20dV%20%3D%20%5Cint%20%5Csigma%20%5Ctimes%20dA)
![Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2](https://tex.z-dn.net/?f=Q%20%3D%20%5Cint%20%5Crho%20%28r%29%20%5C%20dV%20%3D%20%5Csigma%20%5Ctimes4%20%5Cpi%20R%5E2)
∴
![\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2](https://tex.z-dn.net/?f=%5Cint%20%5E%7B2%20%5Cpi%7D_%7B0%7D%20%5Cint%20%5E%7B%5Cpi%7D_%7B0%7D%20%5Cint%20%5E%7BR%7D_%7B0%7D%20%5Crho%20%28r%29%20r%5E2sin%20%5Ctheta%20%20%5C%20dr%20%5C%20d%5Ctheta%20%5C%20d%5Cphi%20%3D%20%5Csigma%20%5Ctimes%204%20%5Cpi%20R%5E2)
![\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2](https://tex.z-dn.net/?f=%5Cint%5E%7B2%20%5Cpi%7D_%7B0%7D%20d%20%5Cphi%2A%20%5Cint%20%5E%7B%5Cpi%7D_%7B0%7D%20%5C%20sin%20%5Ctheta%20d%20%5Ctheta%20%2A%20%5Cint%20%5E%7BR%7D_%7B0%7D%20k%20%5Cdelta%20%28r%20-R%29%20%2A%20r%5E2dr%20%3D%20%5Csigma%20%5Ctimes%204%20%5Cpi%20R%5E2)
![(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2](https://tex.z-dn.net/?f=%282%20%5Cpi%29%282%29%20%2A%20%5Cint%20%5E%7BR%7D_%7B0%7D%20k%20%5Cdelta%20%28r%20-R%29%20%2A%20r%5E2dr%20%3D%20%5Csigma%20%5Ctimes%204%20%5Cpi%20R%5E2)
Thus;
![k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2](https://tex.z-dn.net/?f=k%20%2A%204%20%5Cpi%20%20%5Cint%20%5E%7BR%7D_%7B0%7D%20%20%5Cdelta%20%28r%20-R%29%20%2A%20r%5E2dr%20%3D%20%5Csigma%20%5Ctimes%20%20R%5E2)
![k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2](https://tex.z-dn.net/?f=k%20%2A%20%5Cint%20%5E%7BR%7D_%7B0%7D%20%20%5Cdelta%20%28r%20-R%29%20%20r%5E2dr%20%3D%20%5Csigma%20%5Ctimes%20%20R%5E2)
![k * R^2= \sigma \times R^2](https://tex.z-dn.net/?f=k%20%2A%20R%5E2%3D%20%5Csigma%20%5Ctimes%20%20R%5E2)
![k = R^2 --- (2)](https://tex.z-dn.net/?f=k%20%20%3D%20%20%20R%5E2%20---%20%282%29)
Hence, from equation (1), if k = ![\sigma](https://tex.z-dn.net/?f=%5Csigma)
![\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Crho%20%28r%29%20%3D%20%5Cdelta%2A%20%5Cdelta%20%28r%20-R%29%20%20%5C%20%5C%20%20at%20%20%20%5C%20%5C%20%20%28r%3DR%29%7D)
![\mathbf{\rho (r) =0 \ \ at \ \ rR}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Crho%20%28r%29%20%3D0%20%5C%20%5C%20%20at%20%20%20%5C%20%5C%20%20r%3CR%20%20%5C%20%5C%20%20or%20%5C%20%20%5C%20r%3ER%7D)
To verify the units:
![\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Crho%20%28r%29%20%3D%5Csigma%20%5C%20%2A%20%20%5C%20%5Cdelta%20%28r-R%29%7D)
↓ ↓ ↓
c/m³ c/m³ × 1/m
Thus, the units are verified.
The integrated charge Q
![Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr](https://tex.z-dn.net/?f=Q%20%3D%20%5Cint%20%5Crho%20%28r%29%20%5C%20dV%20%5C%5C%20%5C%5C%20Q%20%3D%20%5Cint%20%5E%7B2%20%5C%20%5Cpi%7D_%7B0%7D%20%5Cint%20%5E%7B%5Cpi%7D_%7B0%7D%20%5Cint%20%5ER_0%20%5Crho%20%28r%29%20%5C%20%5C%20r%5E2%20%5C%20%5C%20%20sin%20%5Ctheta%20%20%5C%20dr%20%5C%20d%5Ctheta%20%5C%20%20d%20%5Cphi%20%20%5C%5C%20%5C%5C%20%20Q%20%3D%20%5Cint%20%5E%7B2%20%5Cpi%7D_%7B0%7D%20%5C%20%20d%20%5Cphi%20%20%5Cint%20%5E%7B%5Cpi%7D_%7B0%7D%20%5C%20sin%20%5Ctheta%20%20%5Cint%20%5ER_%7B0%7D%20%5Crho%20%28r%29%20r%5E2%20%5C%20dr)
![Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr](https://tex.z-dn.net/?f=Q%20%3D%20%282%20%5Cpi%29%20%282%29%20%5Cint%20%5ER_0%20%5Csigma%20%2A%20%5Cdelta%20%28r-R%29%20r%5E2%20%5C%20dr)
![Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr](https://tex.z-dn.net/?f=Q%20%3D%204%20%5Cpi%20%20%5Csigma%20%20%5Cint%20%5ER_0%20%20%2A%20%5Cdelta%20%28r-R%29%20r%5E2%20%5C%20dr)
since ![( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )](https://tex.z-dn.net/?f=%28%20%5Cint%20%5E%7Bxo%7D_%7B0%7D%20%28x%20-x_o%29%20f%28x%29%20%5C%20dx%20%3D%20f%28x_o%29%20%29)
![\mathbf{Q = 4 \pi R^2 \sigma }](https://tex.z-dn.net/?f=%5Cmathbf%7BQ%20%3D%204%20%5Cpi%20R%5E2%20%20%5Csigma%20%20%7D)
Answer:
![7.32\ \text{m/s}](https://tex.z-dn.net/?f=7.32%5C%20%5Ctext%7Bm%2Fs%7D)
Explanation:
= Velocity at initial point = 0
= Pressure in tank = 120 kPa
= Pressure at outlet = 101 kPa
= Density of kerosene = ![750\ \text{kg/m}^3](https://tex.z-dn.net/?f=750%5C%20%5Ctext%7Bkg%2Fm%7D%5E3)
= Tank height = 15 cm
= Height of pipe exit = 0
= Acceleration due to gravity = ![9.81\ \text{m/s}^2](https://tex.z-dn.net/?f=9.81%5C%20%5Ctext%7Bm%2Fs%7D%5E2)
From Bernoulli's equation we have
![\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+Z_2\\\Rightarrow \dfrac{P_1}{\rho g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}\\\Rightarrow v_2=\sqrt{2g(\dfrac{P_1}{\rho g}+Z_1-\dfrac{P_2}{\rho g})}\\\Rightarrow v_2=\sqrt{2\times 9.81(\dfrac{120\times 10^3}{750\times 9.81}+0.15-\dfrac{101\times 10^3}{750\times 9.81})}\\\Rightarrow v_2=7.32\ \text{m/s}](https://tex.z-dn.net/?f=%5Cdfrac%7BP_1%7D%7B%5Crho%20g%7D%2B%5Cdfrac%7Bv_1%5E2%7D%7B2g%7D%2BZ_1%3D%5Cdfrac%7BP_2%7D%7B%5Crho%20g%7D%2B%5Cdfrac%7Bv_2%5E2%7D%7B2g%7D%2BZ_2%5C%5C%5CRightarrow%20%5Cdfrac%7BP_1%7D%7B%5Crho%20g%7D%2BZ_1%3D%5Cdfrac%7BP_2%7D%7B%5Crho%20g%7D%2B%5Cdfrac%7Bv_2%5E2%7D%7B2g%7D%5C%5C%5CRightarrow%20v_2%3D%5Csqrt%7B2g%28%5Cdfrac%7BP_1%7D%7B%5Crho%20g%7D%2BZ_1-%5Cdfrac%7BP_2%7D%7B%5Crho%20g%7D%29%7D%5C%5C%5CRightarrow%20v_2%3D%5Csqrt%7B2%5Ctimes%209.81%28%5Cdfrac%7B120%5Ctimes%2010%5E3%7D%7B750%5Ctimes%209.81%7D%2B0.15-%5Cdfrac%7B101%5Ctimes%2010%5E3%7D%7B750%5Ctimes%209.81%7D%29%7D%5C%5C%5CRightarrow%20v_2%3D7.32%5C%20%5Ctext%7Bm%2Fs%7D)
The exit velocity from the tube is
.