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8_murik_8 [283]
3 years ago
11

Why gravitational force at the center og earth is zero

Physics
2 answers:
KengaRu [80]3 years ago
8 0
F = G Mm/r²
mg = G Mm/r²
g = GM/r²
At centre of earth, r=0
g = GM/0
g =0
Anon25 [30]3 years ago
4 0

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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\huge{\blue{\mathbb{HELLO\:BUDDY}}}

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

❣️ ❣️ Follow me ❣️ Follow me

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

\huge{\green{\tt{\underline{\underline{QUESTION:-}}}}}

Why value of 'g' is zero at center of Earth?

━━━━━━━━━━━━━✦✗✦━━━━━━━━━━━━━━

\huge{\red{\tt{\underline{\underline{EXPLANATION:-}}}}}

When we move towards centre of earth, the mass is equally distributed in all directions.

The mass beneath you = Mass in front of you = Mass behind you

Thus, all the gravitational forces applied cancel each other and acceleration due to gravity (g) at centre of earth on centre of earth becomes zero (0).

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

\huge{\green{\tt{\underline{\underline{PROOF:-}}}}}

The general equation of acceleration due to gravity is;-

g = \frac{GM}{r^2}

Assuming earth to be a perfect sphere and considering its uniform density.

We make the following adjustments in the following equation.

g = \frac{GM}{r^2}

Multiplying and dividing RHS by volume 'V'.

g = \frac{GM}{V} × V × \frac{1}{r^2}

We know that;-

\frac{M}{V} = Density, ρ

Therefore,

g = \frac{ρGV}{r^2}

For sphere;-

V = \frac{4}{3}πr^{3}

This makes

g = \frac{4}{3}πr^{3} × \frac{ρG}{r^2}

g = \frac{4πρG}{3}

So, at center of Earth

since, r = 0

so, g = 0.

━━━━━━━━━━━━━✦✗✦━━━━━━━━━━━━━━

\boxed{\underline{\green{\star{HOPE \: YOU\:GOT \:THE \:ANSWER.}}}}

\boxed{\underline{\pink{\star{MARK \: ME  \: BRAINLIEST.}}}}

\Large{\mathcal{\green{FOLLOW \: ME}}}

Thankyou

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PLEASE HELP
Anvisha [2.4K]

The speed of the rock at 20 m is 34.3 m/s

Explanation:

We can solve this problem by using the law of conservation of energy: the mechanical energy of the rock, sum of its potential energy + its kinetic energy) must be conserved in absence of air resistance. So we can write:

U_i +K_i = U_f + K_f

where :

U_i is the initial potential energy

K_i is the initial kinetic energy

U_f is the final potential energy

K_f is the final kinetic energy

The equation can also be rewritten as  follows:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 100 kg is the mass of the rock

g=9.8 m/s^2 is the acceleration of gravity

h_i = 80 is the initial height

u = 0 is the initial speed  (the rock starts at rest)

h_f = 20 m is the final height of the rock

v is the final speed when h = 20 m

And solving for v, we find:

v=\sqrt{2g(h_i-h_f)}=\sqrt{2(9.8)(80-20)}=34.3 m/s

Learn more about kinetic energy and potential energy here:

brainly.com/question/6536722

brainly.com/question/1198647  

brainly.com/question/10770261  

#LearnwithBrainly

5 0
4 years ago
Select the correct answer.
erica [24]
<span>The correct answer is C:Waves transfer energy, but not matter. A wave does not move matter in the direction of its propagation. It only transfers energy just like the ocean wave traveling many miles away with the water just moving up and down.</span>

6 0
3 years ago
Sasha lifts a couch 8.2 meters from the ground floor of her house to the attic. If the couch has a mass of 120 kg, what is the g
gavmur [86]

As we know that gravitational potential energy is given by

U = mgH

here we have

m = mass = 120 kg

g = 9.81 m/s^2

h = height = 8.2 m

now from above formula

U = 120kg (9.81 m/s^2) (8.2 m)

U = 9653.04 J

so above is the gravitational potential energy of the couch

4 0
3 years ago
Read 2 more answers
a rock is dropped from a height of 80 m and is in free fall what is the velocity of as it reaches the ground 4.0 seconds later
Marta_Voda [28]
Velocity = displacement (distance)/time

v=80m/4s

v=20m/s

velocity = 20 meters per second
8 0
3 years ago
Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0
3 years ago
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