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3 years ago

Why is evaporation of water greater in summer?

2 answers:

7 0

Evaporation of water is greater during the summer, because of the sever change of temperature (Depending where you live) and the increase of light rays on the earths surface, and bodies of water.

prohojiy [21]3 years ago

3 0

More sun greater temperatures <span />

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What are the structures shown below ??

forsale [732]

The body system on the chart

8 0

3 years ago

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

A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if

Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

$F_p = 6*10^{22}N$