Answer:
Work= -7.68×10⁻¹⁴J
Explanation:
Given data
q₁=q₂=1.6×10⁻¹⁹C
r₁=2.00×10⁻¹⁰m
r₂=3.00×10⁻¹⁵m
To find
Work
Solution
The work done on the charge is equal to difference in potential energy
W=ΔU
![Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\ Work=-7.68*10^{-14}J](https://tex.z-dn.net/?f=Work%3DU_%7B1%7D-U_%7B2%7D%5C%5C%20Work%3D-kq_%7B1%7Dq_%7B2%7D%5B%5Cfrac%7B1%7D%7Br_%7B2%7D%7D-%5Cfrac%7B1%7D%7Br_%7B1%7D%7D%20%5D%5C%5CWork%3D%28-9%2A10%5E%7B9%7D%29%2A%281.6%2A10%5E%7B-19%7D%20%29%5E%7B2%7D%5B%5Cfrac%7B1%7D%7B3.0%2A10%5E%7B-15%7D%20%7D-%5Cfrac%7B1%7D%7B2%2A10%5E%7B-10%7D%20%7D%20%5D%5C%5C%20%20Work%3D-7.68%2A10%5E%7B-14%7DJ)
Answer:
an object sliding down hill
Explanation:
On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.
Hello There!
The mantle is the layer located directly under the sima. It is the largest layer of the Earth, 1800 miles thick. The mantle is composed of very hot, dense rock. This layer of rock even flows like asphalt under a heavy weight. This flow is due to great temperature differences from the bottom to the top of the mantle. The movement of the mantle is the reason that the plates of the Earth move! The temperature of the mantle varies from 1600 degrees Fahrenheit at the top to about 4000 degrees Fahrenheit near the bottom!
Hope This helped!
HyperZ ^_^
Answer:
r = 5,085 m
Explanation:
The force exerted by on the surface of the Earth on an electron is its weight
W = F = 9.11 10⁻³¹ 9.8
W = 8.9 10⁻³⁰ N
The electric force between an electron and a proton is given by Coulomb's Law
Fe = k q₁ q₂ / r²
Fe = - k q² / r²
They ask us that W = Fe
W = k q² / r²
r = √ k q² / W
Let's calculate
r = √ 8.99 10⁹ (1.6 10⁻¹⁹)² /8.9 10⁻³⁰
r = √ 25.86
r = 5,085 m
Let's look for the relationship of this distance with the harmonic distance
R / R_atomic = 5,085 / 10⁻¹⁰
R / R_Atomic = 5 10¹⁰
We see that this distance is 10¹⁰ times the interatomic distance, so the gravitational attraction force is very small at atomic scale
F = kq1q2/r<span>2
Where,
F - Coulomb Force
k - constant value which is equal to </span>8.98 × 10^9<span> newton square metre per square coulomb
q1 and q2 - two electric charges
r - distance.
5.8 * 10^5 = 1.5 * 10^-9 / r^2
</span><span>5.8 * 10^5 r^2 = 1.5*10^-9
</span>r^2 = 0.0000258620
r = 0.0050854694
So the distance is equal to 5.09 x 10^-3
