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mezya [45]
3 years ago
7

When neither plate is dense enough to sink into the asthenosphere, the result is a _____ plate boundary.

Physics
1 answer:
hjlf3 years ago
6 0

Answer:  Convergent collision

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C. at a science fair
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Explain the following defects of a simple electric cell:
eimsori [14]

Answer:

Explanation:

The two major defects of simple electric cells causes current supplied to be for short time. These defects are: polarization and local action.

a. Polarization: This is a defect caused by an accumulation of hydrogen bubbles at the positive electrode of the cell. It can be prevented by the use of vent, using a hydrogen absorbing material or the use of a depolarizer.

b. Local Action: This is the gradual wearing away of the electrode due to impurities in the zinc plate. It can be controlled by the amalgamation of the zinc plate before it is used.

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A person on a daily diet of 2,500 calories should get no_more than <br> calories from fat each day.
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35%

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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
HELP PLS
melomori [17]

If the field is in a vacuum, the magnetic field is the dominant factor determining the motion. Since the magnetic force is perpendicular to the direction of travel, a charged particle follows a curved path in a magnetic field. The particle continues to follow this curved path until it forms a complete circle. Another way to look at this is that the magnetic force is always perpendicular to velocity, so that it does no work on the charged particle. The particle’s kinetic energy and speed thus remain constant. The direction of motion is affected but not the speed.

A negatively charged particle moves in the plane of the paper in a region where the magnetic field is perpendicular to the paper (represented by the small × ’s—like the tails of arrows). The magnetic force is perpendicular to the velocity, so velocity changes in direction but not magnitude. The result is uniform circular motion.

4 0
3 years ago
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