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Veseljchak [2.6K]
3 years ago
13

Imagine tying a string to a ball and twirling it around you. How is this similar to the moon orbiting the Earth? In this example

, what is providing the constantly changing, inward force?
Physics
2 answers:
Tema [17]3 years ago
4 0
Its similar to the moon orbiting the earth because lets say that the sing is moon and the ball is earth has the "moon" orbits around the "earth" the string ends up tying around the ball till its no more


i think i hope this example helps you somehow srry that i dont know more then that :/
Archy [21]3 years ago
4 0

Answer:

It is a similar case, given that both can be modeled as uniform circular motion. The inward force for the Earth-moon case is provided by gravity's force, and in the case of the ball-string it is provided by the tension on the string.

Explanation:

There are parellels between both cases:

in the Earth-moon movement, the mass of each astral body generates a pulling force (gravity) which, given the correct circumstances, makes each object revolve a common center of mass, almost in a circular trajectory. Of course, since we are not in such point in space but standing on Earth, we see as the moon is revolving around the earth. The inward force would be gravity in this case. There are other forces involved such as the pulling on the sun, but it is common to both bodies so it is not relevant in this picture.

The ball swirling around you is possible because it is attached to a rope or string, which constantly pulls the ball towards the center (you). The force involved here is called the 'tension' on the rope. The circular movement is a combination of tangential velocity and inward radial force. There are other forces which are more relevant here, such as the friction with the air and the acceleration of gravity of the Earth on the ball, which can actually alter the trajectory of the ball. The pulling must supply some pushing force to counter those forces.

For more information and insight on this problem I recommend videos on uniform circular motion and centripetal forces.

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Dmitry [639]

Answer:

five dollars

Explanation:

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Irina-Kira [14]
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6 0
3 years ago
Read 2 more answers
The distance that a spring will stretch varies directly as the force applied to the spring. A force of 8080 pounds is needed to
xxTIMURxx [149]

Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x   Formula (1)

Where;

F  is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21  inches

Problem development

We replace data in formula 1 to calculate  K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate  F₂

F₂= K*x₂

F₂= (10)*(21)

F₂= 210 pounds

8 0
3 years ago
A rectangular piece of​ cardboard, whose area is 352 square​ centimeters, is made into an open box by cutting a 2​-centimeter sq
Usimov [2.4K]

Answer:

Dimension of cardboard is 22 m by 16 m

Explanation:

Given that,

Area = 352 cm²

Side of each square cutting from corner = 2 cm

Volume of box = 432 cm³

Let the two sides are x and y.

The area of the rectangular piece is

xy=352

y=\dfrac{352}{x} -------- (1)

The volume of the rectangular piece

2(x-4)(\dfrac{352}{x}-4)=432

x^2-38x+352=0

(x-16)(x-22)=0

x=16,22

Put the value of x in the equation (I)

For x = 16

y=\dfrac{352}{16}=22

For x = 22

y=\dfrac{352}{22}=16

Dimension of cardboard is 22 m by 16 m

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4 years ago
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The answer would be C
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