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gayaneshka [121]
2 years ago
8

What would decrease the gravitational pull between the sun and Earth?(1 point)

Physics
1 answer:
VARVARA [1.3K]2 years ago
3 0

The gravitational pull between the sun and Earth decreases if the sun was farther away from the earth.

<h3>What is Gravitational Force?</h3>

Any two bodies' gravitational pull on one another is directly proportional to the product of their masses and inversely proportional to the square of their distance from one another.

How do the planets move in relation to the Sun? The planets' orbits around the Sun are maintained by the gravity of the Sun. The Moon orbits the Earth due to the Earth's gravitational pull on the Moon for the same reason. But if the Earth is being drawn to the Sun, why doesn't it fall into it or the Moon strike the Earth?

The cause of this is because Earth moves at a speed that is perpendicular to the Sun's gravitational pull. The planets are thus kept in their designated orbits by the Sun's attraction as well as their sideways motion. Similar to how the Earth orbits the Moon, the Moon does so without touching it.

F=G(M1M2)/R² where,

F denotes gravitational force between two bodies,

R is the distance between two bodies from their center,

M1 denotes mass of first body and,

M2 denotes mass of second body.

So according to the question, if the distance between two bodies will increase, then the gravitational force between them will also decrease.

Hence, if the sun was farther away from earth, then gravitational pull between the sun and earth will decrease.

To get more information about Gravitational force :

brainly.com/question/12528243

#SPJ1

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On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.70 times as far as he wo
Margarita [4]

Answer:

a)H=84.87 m

b)R=565.27 m

Explanation:

Given that

Speed of ball on earth

U=41.2 m/s

θ=31°

We know that

Maximum height in projectile motion given as

h=\dfrac{U^2sin^2\theta }{2g}

h=\dfrac{41.2^2sin^231}{2\times 9.81}

h=22.94 m

Range in projectile motion given as

r=\dfrac{U^2sin2\theta }{g}

r=\dfrac{41.2^2sin62 }{g}

r=152.77 m

Given that on distance planet ball moves 3.7 times far more as on earth.

So on the distance planet

The maximum height ,H=3.7 h

H= 3.7 x 22.94

H=84.87 m

The range on distance planet

R=3.7 r

R=3.7 x 152.77 m

R=565.27 m

7 0
3 years ago
In golf, a club is swung to hit a small ball resting on the ground. Golfer A holds his club a short distance behind the golf bal
klemol [59]

Answer:

Golfer B hits his golf ball at a higher rate of speed and with more force. This means that golfer B's ball will travel for a longer period of time and a longer distance due to newtons first law, where it says an object in motion stays in motion, with the same speed and in the same direction.

Explanation:

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2 years ago
Describe three common issues that affect societies across places and times
Vesna [10]
Poverty, Natural Disaster, and Disease are 3 answers.
6 0
3 years ago
You place an object 30.0 cm from a concave mirror of 15.0 cm focal length. The object is 1.8 cm tall. Find the image position an
Softa [21]

Answer:30 cm

Explanation:

Given

object distance u=30\ cm

focal length of concave mirror f=15\ cm

height of object h_o=1.8\ cm

Using mirror formula

\Rightarrow \frac{1}{v}+\frac{1}{u}=\frac{1}{f}

\Rightarrow \frac{1}{v}=\frac{1}{15}-\frac{1}{30}

\Rightarrow \frac{1}{v}=\frac{2-1}{30}

\Rightarrow v=30\ cm

and magnification is

\Rightarrow m=\frac{-v}{u}=\frac{h_i}{h_o}

\Rightarrow \frac{-30}{30}=\frac{h_i}{1.8}

\Rightarrow h_i=-1.8\ cm

So height of object is same as of object .

Position :image is formed at the spot where object is placed.

4 0
3 years ago
A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi
Liula [17]

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

<u>a) calculating work done by the force over the journey of the train</u>

F = mx + b  ------ ( 1 )

m = slope  = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8   ----- ( 2 )

hence to determine the work done by the force

W   = \int\limits^7_0 { ( -0.373) x + 2.8  )} \, dx     Note:  the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

<u>b) calculate the speed of the train at the end of its journey</u>

we will apply the work energy theorem

W = 1/2 m*v^2  -  1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 )  ( input values into equation )

 V^2 = 12.11

hence V = 3.48 m/s

6 0
3 years ago
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