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3 years ago

6

Gravitational Potential Energy = mass x gravity x height

2 answers:

Blizzard [7]3 years ago

7 0

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

FinnZ [79.3K]3 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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HELP ASAP!!!! 20 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!

Alexxandr [17]

Answer:

I would have to say the answer is D

Explanation:

because the angle is being changed using the ray box.

8 0

3 years ago

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

8 0

3 years ago

I NEED THIS A.S.A.P I WILL GIVE BRAINLIEST PLEASE HELP!!

VLD [36.1K]

Explanation:

change 0.5 g to kg so 0.005kg then change 100 ml to m so 0.001m so density=mass over volume so from there you can continue

8 0

3 years ago

Read 2 more answers

A ball is dropped off the balcony of a hotel room and it takes 2.8s to fall to the ground . how high above the ground is the bal

RideAnS [48]

The height of the ball above the ground is 38.45 m

First we will calculate the velocity of the ball when it touch the ground by using first equation of motion

v=u+gt

v=0+9.81×2.8

v=27.468 m/s

now the height of the ground can be calculated by the formula

v=√2gh

27.468=√2×9.81×h

h=38.45 m

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3 years ago

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Slav-nsk [51]

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3 years ago