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2 years ago

PLEASE HELP ASAP!! Nitrogen, Helium, and Neon would all be gases at room temperature; Why is this?

1 answer:

7 0

-- The boiling points of the first group are all at temperatures
that are way lower than a comfortable room.
. . . . . Nitrogen . . . -320° F
. . . . . Helium . . . . -452° F
. . . . . Neon . . . . . -411° F

The freezing points of the second group are all at temperatures
that are way higher than a comfortable room.
. . . . . Lithium. . . . . . 357° F
. . . . . Sodium . . . . . 208° F
. . . . . Potassium . . . 146 °F

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Draw the following patterns as they would appear when viewed through a compound microscope.:

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2 years ago

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What is the relation between acceleration due to gravity and radius of the earth?

emmasim [6.3K]

Answer:

As the earth is an oblate spheroid, its radius near the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.

Formula: g = GM/r2

Dimensional Formula: M0L1T-2

Values of g in SI: 9.806 ms-2

Explanation:

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8 0

2 years ago

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Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge

Wittaler [7]

Answer:

F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

r₁₂ = y₁ -y₂

r₁₂ = 0.30 + 0.30

r₁₂ = 0.60 m

let's calculate

F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

r₁₃ = 0.50 m

F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ 0.3 / 0.4

tea = 36.87º

The angle from the positive side of the x-axis is

θ ’= 180 - θ

θ ’= 180 - 36.87

θ ’= 143.13º

sin143.13 = F_13y / F₁₃

F_13y = F₁₃ sin 143.13

F{13y} = 1.697 10⁻² sin 143.13

F_13y = 1.0183 10⁻² N

cos 143.13 = F_13x / F₁₃

F₁₃ₓ = F₁₃ cos 143.13

F₁₃ₓ = 1.697 10⁻² cos 143.13

F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

Fx = F13x + F12x

Fx = -1,357 10-2 +0

Fx = -1.357 10-2 N

Fy = F13y + F12y

Fy = 1.0183 10-2 + 1 10-1

Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

F = Ra (Fx2 + Fy2)

F = RA [(1.357 10-2) 2 + 0.110183 2]

F = 0.111015 N

Let's use trigonometry for the angles

tan tea = Fy / Fx

tea = tan-1 (0.110183 / -0.01357)

tea = 1,448 rad

to find the angle about the positive side of the + x axis

tea '= pi - 1,448

Tea = 1.6936 rad

6 0

3 years ago

Which one should be my pfp?

Zepler [3.9K]

Explanation:

I think the third one coz it's so good

7 0

2 years ago

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A student lefts a bouncy ball to 1.2m, then releases the ball. The ball bounces back up to 0.86m. Construct an explanation as to

Airida [17]

Answer:

the decrease in energy is due to a transformational in internal energy of the body in the rebound.

Explanation:

For this exercise we can calculate the initial and final mechanical energy

Em₀ = U = m g y₁

$Em_{f}$ = U = m g y₂

we look for the variation of the energy

ΔEm = Em_{f} - Em₀

ΔEm = m g (y_{f} -y₀)

ΔEm = m g (0.86 -1.2)

ΔEm = -3.332 m

We can see that there is a decrease in mechanical energy, this is transformed into internal energy of the ball during the impact with the ground, this energy can be formed by several factors such as a part of the friction with the surface, an increase in body temperature or a deformation of the body; there may be a contribution from several of these factors.

In conclusion the decrease in energy is due to a transformational in internal energy of the body in the rebound.

7 0

3 years ago

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