Answer:
The charge on the dust particle is ![q_d = 6.94 *10^{-13} \ C](https://tex.z-dn.net/?f=q_d%20%20%3D%206.94%20%2A10%5E%7B-13%7D%20%5C%20%20C)
Explanation:
From the question we are told that
The length is ![l = 2.0 \ m](https://tex.z-dn.net/?f=l%20%3D%202.0%20%5C%20m)
The width is ![w = 4.0 \ m](https://tex.z-dn.net/?f=w%20%3D%204.0%20%5C%20m)
The charge is ![q = -10\mu C= -10*10^{-6} \ C](https://tex.z-dn.net/?f=q%20%3D%20%20-10%5Cmu%20C%3D%20-10%2A10%5E%7B-6%7D%20%5C%20C)
The mass suspended in mid-air is ![m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg](https://tex.z-dn.net/?f=m_a%20%3D%20%205.0%20%5Cmu%20g%20%3D%20%205.0%20%2A10%5E%7B-6%7D%20%5C%20g%20%3D%20%205.0%20%2A10%5E%7B-9%7D%20%5C%20%20kg)
Generally the electric field on the carpet is mathematically represented as
![E = \frac{q}{ 2 * A * \epsilon _o}](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7Bq%7D%7B%202%20%2A%20%20A%20%20%2A%20%20%5Cepsilon%20_o%7D)
Where
is the permittivity of free space with value ![\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2](https://tex.z-dn.net/?f=%5Cepsilon_o%20%3D%208.85%2A10%5E%7B-12%7D%20%20%5C%20%5C%20%20m%5E%7B-3%7D%20%5Ccdot%20kg%5E%7B-1%7D%5Ccdot%20%20s%5E4%20%5Ccdot%20A%5E2)
substituting values
![E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B-10%2A10%5E%7B-6%7D%7D%7B%202%20%2A%20%20%282%20%2A%204%20%29%20%20%2A%20%208.85%2A10%5E%7B-12%7D%7D)
![E = -70621.5 \ N/C](https://tex.z-dn.net/?f=E%20%3D%20-70621.5%20%5C%20%20N%2FC)
Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles
![F__{E}} = F__{G}}](https://tex.z-dn.net/?f=F__%7BE%7D%7D%20%3D%20%20F__%7BG%7D%7D)
=> ![q_d * E = m * g](https://tex.z-dn.net/?f=q_d%20%2A%20%20E%20%20%3D%20%20m%20%2A%20g)
=> ![q_d = \frac{m * g}{E}](https://tex.z-dn.net/?f=q_d%20%20%3D%20%20%5Cfrac%7Bm%20%2A%20g%7D%7BE%7D)
=> ![q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}](https://tex.z-dn.net/?f=q_d%20%20%3D%20%20%5Cfrac%7B5.0%20%2A10%5E%7B-9%7D%20%2A%209.8%7D%7B70621.5%7D)
=> ![q_d = 6.94 *10^{-13} \ C](https://tex.z-dn.net/?f=q_d%20%20%3D%206.94%20%2A10%5E%7B-13%7D%20%5C%20%20C)
Answer:
A. a material burns out when current is excessive
B. they both involve wave interaction.
Answer:
![\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%20Kinetic%20%5C%20energy%20%5C%20of%20%5C%20the%20%5C%20bear%20%28KE%29%20%3D%2023002.1%20%5C%20J%7D%20)
Given:
Mass of the polar bear (m) = 6.8 kg
Speed of the polar bear (v) = 5.0 m/s
To Find:
Kinetic energy of the polar bear (KE)
Explanation:
Formula:
![\boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Cbold%7B%5Csf%20KE%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20m%20%7Bv%7D%5E%7B2%7D%20%7D%7D)
Substituting values of m & v in the equation:
![\sf \implies KE = \frac{1}{2} \times 380.2 \times {11}^{2}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20KE%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%20380.2%20%5Ctimes%20%20%7B11%7D%5E%7B2%7D%20)
![\sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 190.1 \times 121](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20KE%20%3D%20%5Cfrac%7B1%7D%7B%20%5Ccancel%7B2%7D%7D%20%20%5Ctimes%20%20%5Ccancel%7B2%7D%20%5Ctimes%20190.1%20%5Ctimes%20121%20)
![\sf \implies KE = 190.1 \times 121](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20KE%20%3D%20190.1%20%5Ctimes%20121%20)
![\sf \implies KE = 23002.1 \: J](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20KE%20%3D%2023002.1%20%5C%3A%20J)
![\therefore](https://tex.z-dn.net/?f=%20%5Ctherefore%20)
Kinetic energy of the polar bear (KE) = 23002.1 J