Weight is measured as the downward force of _________

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago

On a violin, the highest note Mandy can play is an A-note, which produces a sound wave with a high frequency. The lowest note sh

lora16 [44]

I think the answer would be: The G-note's wavelength is longer

Here are the formula to calculate wavelength

Wavelength = Wave speed/Frequency

Which indicates that the wavelength will become larger as the frequency became smaller.

6 0

3 years ago

Read 2 more answers

Matching

Anon25 [30]

Answer:

D. Principle of original horizontality

B. Principle of faunal succession

A. Uniformitarianism

C. Principle of superposition

Explanation:

Question 1

The principle of original horizontality is one of the foremost relative dating principle that is wide used in stratigraphy.

It states that "sedimentary rocks are laid down flatly on top one another in a sedimentary basin".

Sedimentary rocks will only vary vertically, but laterally, they are uniform and internally homogeneous in space. This is why most sedimentary rocks are stratified and laid layers upon layers just like the pages of a book.

Each layer is called a bed and often times are laterally continuous in space within the same basin.

Whenever we see beds not horizontally continuous, we can conclude that a tectonic event must have disrupted the sequence and it came after the it was formed.

Question 2

Principle of faunal succession succession was proposed by Williams Smith, an English Geologist and a canal worker in the 19th century.

Based on this principle, sedimentary rocks can placed in their proper chronostratigrahic framework based on the fossils they contain in them.

This principle is hinged on theory of evolution.
It is widely accepted that organisms evolved from one another.
Rocks often bear these records in fossil remains and this can help us appropriately fit rocks to the time they were formed.

Question 3

The principle of uniformitarianism was one of the disruptive proposition in earth science.

A Scottish name James Hutton while in the country side made this proposition as he observed how landform in his native changed with each episode of season.

The principle proposes that "the processes occurring today have occurred in the time past at the same rate".

This way, it was much more easier to understand how land changes in pre-historic times have occurred.

Before his theory, the principle of catastrophism was the widely accepted one. This theory suggested that events occurred rapidly and changes to the surface are much more faster.

Question 4

The principle of superposition is one of the relative dating principles. It proposes that "in an undeformed land sequence, the oldest rock is at the bottom and the youngest on top".

The first sediment to get deposited fills the bottom as it aggregates upward. This leaves the youngest lithology to the top of strata.

The principle is correct for undeformed or undisturbed rock strata.

Where the sediments are disturbed, the formation might be overturned and this principle might be difficult to apply.

3 0

3 years ago

Is a truck braking to avoid an accident moving at constant

frutty [35]

Answer:

its not moving at a constant velocity because it is slowing down

Explanation:

3 0

3 years ago

Read 2 more answers

A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg

Dahasolnce [82]

Answer:

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂

10 m/s = 5 v₁ + 4 v₂

Assuming an elastic collision, kinetic energy is conserved.

½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²

m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²

(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²

280 m²/s² = 5 v₁² + 4 v₂²

Substituting:

v₂ = (10 − 5 v₁) / 4

280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²

280 = 5 v₁² + (10 − 5 v₁)² / 4

1120 = 20 v₁² + (10 − 5 v₁)²

1120 = 20 v₁² + 100 − 100 v₁ + 25 v₁²

0 = 45 v₁² − 100 v₁ − 1020

0 = 9 v₁² − 20 v₁ − 204

0 = (9 v₁ + 34) (v₁ − 6)

v₁ = -3.78 m/s or 6 m/s

u₁ = 6 m/s, so v₁ = -3.78 m/s. Solving for v₂:

v₂ = (10 − 5 v₁) / 4

v₂ = 7.22 m/s

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

6 0

3 years ago

Weight is measured as the downward force of __________.