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iris [78.8K]
3 years ago
14

What is the most common state of matter in the universe

Physics
1 answer:
ch4aika [34]3 years ago
7 0

Answer:

<u><em>Plasma</em></u>

Explanation:

<u><em>Plasma</em></u> is the most common because plasma is a gas that has been energized to the point that some of the electrons break

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A lad, waiting for his friend walks in the sidewalk, in front of her house, from the front door, first, he moves towards the Pos
Andreas93 [3]

His total displacement from his original position is -1 m

We know that total displacement of an object from a position x to a position x', d = final position - initial position.

d = x' - x

If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.

Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.

His final position, x" after moving back 6 m is gotten from

x" - x' = -6 m

x" = -6 + x'

x" = -6 + 5

x" = -1 m

Thus, his total displacement from his original position is

d = final position - initial position

d = x" - x

d = -1 m - 0 m

d = -1 m

So, his total displacement from his original position is -1 m

Learn more about displacement here:

brainly.com/question/17587058

3 0
3 years ago
Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete
Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m      

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(15)^2}{22.5}

a=10\ m/s^2

We know that, g=9.8\ m/s^2

a=\dfrac{10\times g}{9.8}

a=1.02\ g

Hence, this is the required solution.                                              

5 0
3 years ago
determine the force of gravitational attraction between a 78kg boy sitting 2 meters away from a 65kg girl.
bixtya [17]

Answer:

F=8.45\times 10^{-8}\ N

Explanation:

Given that,

Mass of a boy is 78 kg

Mass of a girl is 65 kg

We need to find the force of gravitational attraction between them if they are 2 m away.

The formula for the gravitational force is given by :

F=\dfrac{Gm_1m_2}{r^2}\\\\F=\dfrac{6.67\times 10^{-11}\times 78\times 65}{(2)^2}\\\\=8.45\times 10^{-8}\ N

So, the force between them is 8.45\times 10^{-8}\ N.

8 0
3 years ago
Dish-shaped reflectors are used to steer microwaves in order to establish communications links between nearby buildings. Those r
Ad libitum [116K]

A "screen" or even just a set of parallel bars are highly reflective to electromagnetic waves as long as the open spaces are small compared to the wavelengths.

"Grid" dishes work fine ... with less weight and less wind resistance ... for frequencies below about 3 GHz. (Wavelengths of at least 10 cm.)

(I even worked on a microwave system in South America where huge grid dishes were used on a 90-mile link.)

5 0
2 years ago
Prove(show) ''T=2π√(l/g)''​
Nonamiya [84]

Answer:

Time period for Simple pendulum, T=2\pi\sqrt{\frac{l}{g}

Explanation:

The Simple Pendulum

Consider a small bob of mass m is tied to extensible string of length l that is fixed to rigid support. The bob is oscillating in the plane about verticle.

       Let \theta is the angle made by string with vertical  during oscillation.

Vertical component of the force on bob, F=-mg\sin\theta

Negative sign shows that its opposing the motion of bob.

Taking \theta as very small angle then, \sin\theta\sim\theta

F=-mg\theta    

Let x is the displacement made by bob from its mean position ,

then, \theta=\frac{x}{l}

so, F=-mg\frac{x}{l}                ........(1)

Since, pendulum is in hormonic motion,

as we know, F=-kx

where k is the constant and k=m\omega^{2}

F=-m\omega^2x                   .........(2)

From equation (1) and (2)

-m\omega^2x=-mg\frac{x}{l}

\omega=\sqrt{\frac{g}{l}}

Since, \omega=\frac{2\pi}{T}

\frac{2\pi}{T}=\sqrt{\frac{g}{l}

T=2\pi\sqrt{\frac{l}{g}}

6 0
3 years ago
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