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White raven [17]
3 years ago
13

Michael has a substance that he puts in Container 1. The substance has a volume of 5 cubic meters. He then puts the substance in

Container 2, and it now has a volume of 10 cubic meters. In which phase
is the substance?
Physics
1 answer:
m_a_m_a [10]3 years ago
3 0
Gas because liquids and solids volumes don't change from switching containers.
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If velocity is 0.2m/s, what is the kinetic energy?
Natalija [7]

\huge\boxed{♧ \: \mathfrak{ \underline{Answer} \: ♧}}

we know,

\boxed{kinetic  \:  \: energy =  \frac{1}{2} m {v}^{2} }

So,

\longmapsto \dfrac{1}{2}  \times 10 \times 0.2 \times 0.2

\longmapsto0.2 \: joules

3 0
4 years ago
How do you find the rest mass (kg) of a 3.1 eV electron?
Scilla [17]

Answer:

Explanation:

The rest energy of any substance is defined by the Einstein's mass energy equivalence relation. Thus the rest mass of a electron is 9.11x10^-31 kg. The speed of light is 299,792,458 m/s. Thus multiplying the square of speed of light with the rest mass of electron gives the rest energy of the electron.

3 0
3 years ago
Fig.4.1
Nutka1998 [239]

Answer:

A

Explanation:

7 0
4 years ago
what is the ratio of ey,final, the maximum value of the y-component of the electric field immediately following the last polariz
OleMash [197]

That ratio is 2 .

<h3>What is ratio?</h3>

A ratio is the  comparison of the  two numbers bydivision.

Taking the first two outputs, or the  -1/8 and -1/4, we can divide not the second one by the first one to find the

ratio:

-1/4-1/8

When dividing fractions, we multiply by the reciprocal:

-1/4x-8/1

To multiply fractions, multiply straight across:

(-1x-8)/(4x1) 8/4=2

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brainly.com/question/25927869

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8 0
1 year ago
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0
3 years ago
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