Describe what happens to the gravitational force when the Sun, Earth and Moon are in a line.

Nata [24]

Answer:

Force can also be described intuitively as a push or a pull. ... It is measured in the SI unit of newtons and represented by the symbol F. The original form of Newton's second law states that the net force acting upon an object is equal to the rate at which its momentum changes with time.

Explanation:

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6 0

3 years ago

irius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Ano

Norma-Jean [14]

The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

p is the parallax angle observed

Here for Regulus, we get:

S = 1 / p

= 1 / (0.042) [ 1 parsecs = 1 arcseconds ]

= 23.81 parsecs

We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

Learn more about astronomical units here:

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6 0

2 years ago

The spherical annulus of cometary nuclei surrounding the solar system is called the ___.

Bingel [31]

Answer:

Oort Cloud

Explanation:

Hundreds of millions of comet nuclei forming a spherical region that surrounds the solar system is called Oort cloud.

The Oort cloud revolves at tens of thousands of the distance of the earth from the sun.

Comets are the condensed water and dust particles that orbit around the sun with inward helix leaving behind a tail of vapours and vaporize completely as they get closer to the sun during their revolution around it.

They have a nebulous appearance and extremely elongated orbits.

3 0

3 years ago

The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these

Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

$r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\$

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

$r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\$

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

$F = \frac{kq_{1}q_{2}}{r^2}\\$

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

$F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\$

F = 3.86 x 10⁻⁶ N

5 0

3 years ago

In a 49 s interval, 595 hailstones strike a glass window of an area of 0.954 m at an angle of 25° to the window surface. Each ha

eduard

Average force on the window: 0.32 N

Explanation:

The average force exerted on the window is given by Newton's second law

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the net change in momentum of the hailstones in a time interval of $\Delta t$

In order to find the change in momentum, we have to consider only the component of the hailstone's momentum perpendicular to the window, therefore:

$p_i =m u sin \theta$ is the initial momentum of one hailstone, with

m = 7 g = 0.007 kg is the mass

$u=4.5 m/s$ is the initial speed

$\theta=25^{\circ}$ is the angle with the window

The final momentum is

$p_f = mv sin \theta$

where

v = 4.5 m/s is the final speed (the collision is elastic so the speed is equal, while the direction changes)

$\theta=-25^{\circ}$ (after the rebound, the direction has changed)

So the change in momentum of 1 hailstone is

$\Delta p = mv sin(-25^{\circ})-mu sin(25^{\circ})=-2mu sin(25^{\circ})=-0.0266 kg m/s$

We are interested only in the magnitude, so

$\Delta p = 0.0266 kg m/s$

There are 595 hailstones hitting the window in 49 s, so the total change in momentum is

$\Delta p = 595\cdot 0.0266 = 15.8 kg m/s$

And therefore, the average force on the window is

$F=\frac{\Delta p}{\Delta t}=\frac{15.8}{49}=0.32 N$

Learn more about force:

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3 0

3 years ago