Answer:
Electric Field = E = 36.848 N/C
Explanation:
In accordance with Columb's law
E = k Q1 Q2 / r.r = 8.99 x 10^9 x 5.0 x 10^-6 x 5.8 x 10^-6 / 0.084 x 0.084
= 36948.6961 x 10^-3 = 36.848 N/C
Answer:
Δ h = 52.78 m
Explanation:
given,
Atmospheric pressure at the top of building = 97.6 kPa
Atmospheric pressure at the bottom of building = 98.2 kPa
Density of air = 1.16 kg/m³
acceleration due to gravity, g = 9.8 m/s²
height of the building = ?
We know,
Δ P = ρ g Δ h
(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h
11.368 Δ h = 600
Δ h = 52.78 m
Hence, the height of the building is equal to 52.78 m.
Answer:
Electric potential energy at the negative terminal: ![1.92\cdot 10^{-18}J](https://tex.z-dn.net/?f=1.92%5Ccdot%2010%5E%7B-18%7DJ)
Explanation:
When a particle with charge
travels across a potential difference
, then its change in electric potential energy is
![\Delta U = q \Delta V](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20q%20%5CDelta%20V)
In this problem, we know that:
The particle is an electron, so its charge is
![q=-1.60\cdot 10^{-19}C](https://tex.z-dn.net/?f=q%3D-1.60%5Ccdot%2010%5E%7B-19%7DC)
We also know that the positive terminal is at potential
![V_+=0V](https://tex.z-dn.net/?f=V_%2B%3D0V)
While the negative terminal is at potential
![V_-=-12 V](https://tex.z-dn.net/?f=V_-%3D-12%20V)
Therefore, the potential difference (final minus initial) is
![\Delta V = -12-0 = -12 V](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20-12-0%20%3D%20-12%20V)
So, the change in potential energy of the electron is
![\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20%28-1.6%5Ccdot%2010%5E%7B-19%7D%29%28-12%29%3D1.92%5Ccdot%2010%5E%7B-18%7DJ)
This means that the electron when it is at the negative terminal has
of energy more than when it is at the positive terminal.
Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is
![1.92\cdot 10^{-18}J](https://tex.z-dn.net/?f=1.92%5Ccdot%2010%5E%7B-18%7DJ)
Explanation:
Given that,
Initial speed of the sports car, u = 80 km/h = 22.22 m/s
Final speed of the runner, v = 0
Distance covered by the sports car, d = 80 km = 80000 m
Let a is the acceleration of the sports car. It can be calculated using third equation of motion as :
![v^2-u^2=2ad](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2ad)
![a=\dfrac{v^2-u^2}{2d}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2d%7D)
![a=\dfrac{0-(22.22)^2}{2\times 80000}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B0-%2822.22%29%5E2%7D%7B2%5Ctimes%2080000%7D)
![a=-0.00308\ m/s^2](https://tex.z-dn.net/?f=a%3D-0.00308%5C%20m%2Fs%5E2)
Value of g, ![g=9.8\ m/s^2](https://tex.z-dn.net/?f=g%3D9.8%5C%20m%2Fs%5E2)
![a=\dfrac{-0.00308}{9.8}\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B-0.00308%7D%7B9.8%7D%5C%20m%2Fs%5E2)
![a=(-0.000314)\ g\ m/s^2](https://tex.z-dn.net/?f=a%3D%28-0.000314%29%5C%20g%5C%20m%2Fs%5E2)
Hence, this is required solution.
Speed = frequency * wavelength