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2 years ago

9

Since the universe is infinite, the probability of events and reoccurrence is infinite. So that means that there is a chance, if

not 100% chance, that SOMEWHERE there is another earth, with another you, living the same life, at the same time. What are your thoughts about this??? that's so crazy to me

1 answer:

il63 [147K]2 years ago

7 0

Answer:

That insane it might be true because a planet sometimes quoted to be an Earth 2.0 or Earth's Cousin based on its characteristics; also known by its Kepler Object of Interest designation KOI-7016.01) is an exoplanet orbiting the Sun-like star Kepler-452 about 1,402 light-years (430 pc) from Earth in the constellation Cygnus.

Explanation:

You might be interested in

Examples of applied force

Vladimir [108]

Answer:

Push - The most common form of force is a push through physical contact (like a lawnmower or shopping cart)

Pull - You can apply a force by directly pulling on an object (like pulling a wagon)

Explanation:

4 0

3 years ago

Read 2 more answers

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0

2 years ago

. A 2.0-kg block is on a perfectly smooth ramp that makes an angle of 30° with the horizontal. (a) What is the block’s accelerat

AfilCa [17]

Answer:

a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N

Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

sin 30 = Wx / W

cos 30 = Wy / W

Wx = W sin30

Wy = W cos 30

Let's write the equations on each axis

X axis

Wx = ma

Y Axis

N- Wy = 0

N = Wy = mg cos 30

N = 2.0 9.8 cos 30

N = 16.97 N

We calculate the acceleration

a = Wx / m

a = mg sin 30 / m

a = g sin 30

a =9.8 sin 30

a = 4.9 m / s²

b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

F -Wx = 0

F = Wx

F = m g sin 30

F = 2.0 9.8 sin 30

F = 9.8 N

5 0

3 years ago

An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver

Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, $E=2.3\times 10^{-10}\ N/C$

Vertical distance covered, $s=1\ \mu m=10^{-6}\ m$

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

$v^2-u^2=2as$

$v^2=2as$ .............(1)

Electric force is $F_e$ and force of gravity is $F_g$ . As both forces are acting in downward direction. So, total force is:

$F=mg+qE$

$F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}$

$F=4.57\times 10^{-29}\ N$

Acceleration of the electron, $a=\dfrac{F}{m}$

$a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}$

$a=50.21\ m/s^2$

Put the value of a in equation (1) as :

$v=\sqrt{2as}$

$v=\sqrt{2\times 50.21\times 10^{-6}}$

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0

3 years ago

A penny has a mass of 2.50 g, a diameter of 19.55 mm, and a thickness of 1.55 mm. Calculate the density of the material from whi

Dmitry [639]

Density = (mass) divided by (volume)

We know the mass (2.5 g). We need to find the volume.

The penny is a very short cylinder.
The volume of a cylinder is (π · radius² · height).
The penny's radius is 1/2 of its diameter = 9.775 mm.
The 'height' of the cylinder is the penny's thickness = 1.55 mm.

Volume = (π) (9.775 mm)² (1.55 mm)

   = (π) (95.55 mm²) (1.55 mm)

   = (π) (148.1 mm³)

   = 465.3 mm³

We know the volume now. So we could state the density of the penny,
but nobody will understand what we have. Here it is:

   mass/volume = 2.5 g / 465.3 mm³ = 0.0054 g/mm³ .

Nobody every talks about density in units of ' gram/(millimeter)³ ' .
It's always ' gram / (centimeter)³ '.
So we have to convert our number for the volume.

   (0.0054 g/mm³) x (10 mm / cm)³

   = (0.0054 x 1,000) g/cm³

   = 5.37 g/cm³ .

This isn't actually very close to what the US mint says for the density
of a penny, but it's in a much better ball park than 0.0054 was.

4 0

3 years ago