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2 years ago

If gravity between the Sun and Earth suddenly vanished, Earth would continue moving in

Physics

1 answer:

Ksenya-84 [330]2 years ago

6 0

Answer:

Earth would continue moving by uniform motion, with constant velocity, in a straight line

Explanation:

The question can be answered by using Newton's first law of motion, also known as law of inertia, which states that:

"an object keeps its state of rest or of uniform motion in a straight line unless acted upon by an external net force different from zero"

This means that if there are no forces acting on an object, the object stays at rest (if it was not moving previously) or it continues moving with same velocity (if it was already moving) in a straight line.

In this problem, the Earth is initially moving around the Sun, with a certain tangential velocity v. When the Sun disappears, the force of gravity that was keeping the Earth in circular motion disappears too: therefore, there are no more forces acting on the Earth, and so by the 1st law of Newton, the Earth will continue moving with same velocity v in a straight line.

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If the 5-N force and the 12-N force form a 90 degree angle, what is the magnitude of the force acting in the direction of the da

Leni [432]

Answer:

Explanation:

<em>Kindly see attached file for your reference</em>

Step one:

given data

the horizontal component of the force= 12N

the vertical component of the force= 5N

The dashed arrow represents the hypotenuse of the triangle, hence the resultant of the force system.

By implication of this, we will use the Pythagoras theorem to solve for the resultant force

Step two:

$F_R=\sqrt{F_H^2+F_V^2}\\\\F_R= \sqrt{12^2+5^2}\\\\F_R=\sqrt{144+25}\\\\F_R=\sqrt{169}\\\\F_R=13N$

3 0

1 year ago

Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?

Oksi-84 [34.3K]

Answer:

$2.32\times 10^{-11}$

Explanation:

First number is $4.48\times 10^{-8}$

Second number is $5.2\times 10^{-4}$

We need to multiply the two numbers.

$4.48\times 10^{-8}\times 5.2\times 10^{-4}=(4.48\times 5.2)\times 10^{(-8-4)}\\\\=23.296\times 10^{-12}$

In scientific notation : $2.32\times 10^{-11}$

Hence, this is the required solution.

8 0

2 years ago

A reflecting telescope is built with a 20-cm-diameter mirror having a 1.00 m focal length. it is used with a 10× eyepiece.

kondaur [170]

I really wish I could be helping you. I don't know.

3 0

2 years ago

Two parallel co-axial disks are floating in deep space (far from sun and planets). Each disk is 1 meter in diameter and the disk

HACTEHA [7]

Answer:

T₂ = 5646 K

Explanation:

Let's start by finding the power received by the first disc, for this we use Stefan's law

P = σ. A e T⁴

Where next is the Stefam-Bolztmann constant with value 5,670 10-8 W / m² K⁴, A is the area of the disk, T the absolute temperature and e the emissivity that for a black body is 1

The intensity is defined as the amount of radiation that arrives per unit area. For this we assume that the radiation expands uniformly in all directions, the intensity is

I = P / A

Writing this expression for both discs

I₁ A₁ = I₂ A₂

I₂ = I₁ A₁ / A₂

The area of a sphere is

A = 4π r²

I₂ = I₁ (r₁ / r₂)²

r₂ = r₁ ± 5

I₁ = I₂ ( (r₁ ± 5)/r₁)²

Let's write the Stefan equation

P / A = σ e T⁴

I = σ e T⁴

This is the intensity that affects the disk, substitute in the intensity equation

σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²

The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white

e₁ T₁⁴ = T₂⁴ (r1 + 5)²/r₁²

T₁ = T₂ {(e₂/e₁)}^{1/4} √(1 ± 1/ r₁)

If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term

(1 ±x) n = 1 ± n x

Where x = 5 / r₁ << 1

We replace

T₁ = T₂ {(e₂/e₁)}^{1/4} (1 ± ½ 5/r1)

T₁ = T₂ {(e₂)}^{1/4} (1 ± 5/2 1/r1)

If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach

T₁ = T₂ {(e₂)}^{1/4} ¼

T<u>₁</u> = T₂ 0.9 $0.9^{1/4}$