Given Information:
Radius = ra = 2.60 cm = 0.026 m
Density = J = 15.0 nC/m
change in potential difference = ΔV = 200 V
Required Information:
Distance = d = ?
Answer:
distance = 0.088 m
Explanation:
As we know
ΔV = Vb - Va = J/4πε₀*ln(rb/ra)
Where ra and rb is the point where potential difference is Va and Vb respectively
1/4πε₀ = 9x10⁹ N.m²/C²
We want to find the distance d = rb - ra
ΔV = J/4πε₀*ln(rb/ra)
200 = 9x10⁹*15x10⁻⁹*ln(rb/ra)
200/135 = ln(rb/ra)
1.48 = ln(rb/ra)
taking e on both sides yields
e^(1.48) = rb/ra
4.39 = rb/ra
rb = 4.39*0.026
rb = 0.114 m
Therefore, the required distance is
d = rb - ra
d = 0.114 - 0.026
d = 0.088 m
Therefore, the other probe must be placed 0.088 m from the surface so that the voltmeter reads 200 V
Answer:
Explanation:
Force of gravity = GMm/r^2 = ma
a being the acceleration due to gravity at some distance r from the center of the Earth. I'll use 6400 km for the radius of the Earth so r = 6734 km or 6734000 meters
a = GM/r^2
plugging in G = 6.67 x 10^-11
M is the mass of the Earth = 6 x 10^24
and r is from above
a = 8.825 m/s^2 = 0.9g
so 90% the acceleration of gravity on the surface.
Answer:
Explanation:
Given data:
mass of block is 
radius of block = 0.061 m
moment of inertia is 
D is distance covered by block = 0.65 m
speed of block is 1.705 m/s
From conservation of momentum we have
![0.84 \times 9.81 \times 0.65 = \frac{1}{2}\times 0.84 \times 1.705^2 +\frac{1}{2} \times 6.2 \times 10^{-3} [\frac{1.705}{0.061}]^2 + E_l](https://tex.z-dn.net/?f=0.84%20%5Ctimes%209.81%20%5Ctimes%200.65%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%200.84%20%5Ctimes%201.705%5E2%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%206.2%20%5Ctimes%2010%5E%7B-3%7D%20%5B%5Cfrac%7B1.705%7D%7B0.061%7D%5D%5E2%20%2B%20E_l)
solving for energy loss
Explanation:
This question is not feasible. There is no way to calculate the energy needed because the question is missing the final temperature
