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Tasya [4]
3 years ago
6

In which state of matter do the particles spread apart and fill all the space available to them?

Physics
2 answers:
Sav [38]3 years ago
7 0
This is a gas or C!!!!!!!!!!!!!!!!!
Alex_Xolod [135]3 years ago
3 0
Answer choice C. Gas is your answer
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A 2.43ug particle moves at 1.97 x 108 m/s. What is its momentum?
Ira Lisetskai [31]

Answer:

0.48 kgm/s

Explanation:

m = mass of the particle = 2.43 μg = 2.43 x 10⁻⁶ x 10⁻³ kg = 2.43 x 10⁻⁹ kg

v  = velocity of the particle = 1.97 x 10⁸ m/s

p = momentum of the particle

momentum of the particle is given as

p = m v

inserting the values

p = (2.43\times 10^{-9})(1.97\times 10^{8})

p = 0.48 kgm/s

4 0
3 years ago
Which of these statements is most likely correct about the Nebular theory of formation of planets?
exis [7]
Correct answer is D.

Explanation:
A) is not correct answer because this type of theory can not become law. Laws are properties that are same in any part of universe. Nebular theory is not correct for every part of universe.

B) is not correct answer because this theory could be replaced if some evidence show that some other theory is more likely to be correct.

C) is not correct answer because the study has been done on other nebulas in our galaxy. There are many nebulas and by obserwing them this theory was developed.
4 0
3 years ago
A ball is thrown so that its initial vertical and horizontal components of velocity are 30 m/s and 15 m/s, respectively. Estimat
mihalych1998 [28]

Answer:

H = 45 m

Explanation:

First we find the launch velocity of the ball by using the following formula:

v₀ = √(v₀ₓ² + v₀y²)

where,

v₀ = launching velocity = ?

v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s

v₀y = Vertical Component of Launch Velocity = 30 m/s

Therefore,

v₀ = √[(15 m/s)² + (30 m/s)²]

v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

θ = tan⁻¹ (v₀y/v₀ₓ)

θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

H = (v₀² Sin² θ)/2g

H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)

<u>H = 45 m</u>

6 0
3 years ago
An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
3 years ago
Which letter (A, B, or C) shows where you should apply the most effort to lift the stone?
sergij07 [2.7K]
A becuz its at da it dont got no wa
8 0
3 years ago
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