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4 years ago

Gravitational Potential Energy = mass x gravity x height

Physics

2 answers:

Blizzard [7]4 years ago

7 0

M= gpe / gh
G= gpe / mh
H=gpe / mg

FinnZ [79.3K]4 years ago

7 0

Explanation:

The mathematical expression for the gravitational potential energy (GPE) is given by :

$GPE=m\times g\times h$

Where

m is the mass of an object

g is the acceleration due to gravity

h is the height

(a) Solve for m :

$m=\dfrac{GPE}{gh}$

(b) Solve for g :

$g=\dfrac{GPE}{mh}$

(c) Solve for h :

$h=\dfrac{GPE}{mh}$

Hence, this is the required solution.

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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field

Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$

Therefore, the electric field strength at the mid-point between the two rings is zero.

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3 years ago

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Answer:

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3 years ago