To solve this problem we will apply the concepts related to energy conservation. Here we will understand that the potential energy accumulated on the object is equal to the work it has. Therefore the relationship that will allow us to calculate the height will be


Here,
m = mass
g = Acceleration due to gravity
h = Height
our values are,



Replacing,



Then the height is 32.83m.
A.) For letter a, we use the law of universal gravitation using the constant G = 6.674×10−<span>11 m3</span>⋅kg−1⋅s−<span>2
Grav. F = G*m1*m2*(1/d^2)
m1 is mass of electron = </span>9.11 × 10-31<span> kg
m2 is mass of proton = </span>1.67 × 10<span>-27 kg
d = 4.5 nm = 4.5 x 10^-9 m
Grav F = 5.01 x 10^-51 N
b.) </span>For letter b, we use the Coulomb's using the constant k = 9×10^9 N
Electric force = k*Q1*Q2*(1/d^2)
Q1 is charge of electron = -1.6 × 10-19 C
Q2 is charge of proton = +1.6 × 10-19 C
Electric force = 1.14 x 10^-11 N
Answer:
d' = d /2
Explanation:
Given that
Distance = d
Voltage =V
We know that energy in capacitor given as



If energy become double U' = 2 U then d'



2 d ' = d
d' = d /2
So the distance between plates will be half on initial distance.
Answer:
Vi = 24.14 m/s
Explanation:
If we apply Law of Conservation of Energy or Work-Energy Principle here, we get: (neglecting friction)
Loss in K.E of the Rock = Gain in P.E of the Rock
(1/2)(m)(Vi² - Vf²) = mgh
Vi² - Vf² = 2gh
Vi² = Vf² + 2gh
Vi = √(Vf² + 2gh)
where,
Vi = Rock's Speed as it left the ground = ?
Vf = Final Speed = 17 m/s
g = 9.8 m/s²
h = height of rock = 15 m
Therefore,
Vi = √[(17 m/s)² + 2(9.8 m/s²)(15 m)]
Vi = √583 m²/s²
<u>Vi = 24.14 m/s</u>
Answer:
5.33 cm
Explanation:
The lens equation states that:

where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem,
p = 8 cm
q = 16 cm ( the sign is positive since the image is real, which means it is formed on the other side of the lens)
Substituting into the equation,

