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2 years ago

PLEASE HELP ASAP

Physics

1 answer:

77julia77 [94]2 years ago

8 0

Answer:

<em>The first choice (32m/s) is the closest to the answer</em>

Explanation:

The magnitude of a vector is the distance between the initial and the end point of the vector.

Being Vx and Vy the horizontal and vertical components of the vector V respectively, the magnitude of V is calculated as:

$\mid \vec{V} \mid =\sqrt{V_x^2+V_y^2}$

The components of the velocity of the physics student's projectile launcher are Vx=28 m/s and Vy=15 m/s.

Calculate the magnitude of the velocity:

$\mid \vec{V} \mid =\sqrt{28^2+15^2}$

$\mid \vec{V} \mid =\sqrt{784+225}$

$\mid \vec{V} \mid =\sqrt{1009}$

$\mid \vec{V} \mid =31.8 \ m/s$

The first choice (32m/s) is the closest to the answer

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Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

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