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ValentinkaMS [17]
2 years ago
11

At what distance is the electrostatic force between two protons equal to the weight of one?.

Physics
1 answer:
erik [133]2 years ago
8 0

0.118 m is the distance between the two protons.

Mass of proton = 1.6726 × 10⁻²⁷ kg

Weight of proton= 1.6726 × 10⁻²⁷ x 9.81 N

                            = 1.6408 × 10⁻²⁶ N

Charge of proton = 1.602 × 10⁻²⁹ C

The force between two protons = kq²/r²     where, K is a proportionality    

                                                            constant, q is a charge of proton and  

                                                         r is the distance between two protons.          

                                                       = 9 × 10⁹ × (1.602×10⁻¹⁹)²/r²      

To calculate distance :

  Weight of proton= Force between protons

 ⇒   1.6408 × 10⁻²⁶ N = 9 × 10⁹ × (1.602×10⁻¹⁹)²/r²

 ⇒    r = 0.118m

Therefore, 0.118 m is the distance between the two protons.

Learn more about electrostatic force here:

brainly.com/question/18108470

#SPJ4

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Answer:

∅ = 89.44°

Explanation:

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Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = V_{1x}tcos ∅     .......................... Equation 1

where V_{1x} = velocity in the X - direction

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Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

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Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

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  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

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              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6Q^{2} + 19.6

Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

                    Q_{1} = 101.6291

                      Q_{2} = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

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Answer

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