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alisha [4.7K]
3 years ago
5

Need help on these 2 questions please????? Hurry

Physics
1 answer:
fredd [130]3 years ago
3 0

Answer:

Question 1 the anwser is 300 meters.  Question 2 the answer is speed.

is Explanation:

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
A crate remains stationary after it has been placed on a ramp inclined at an angle with the horizontal. Which of the following s
Ganezh [65]

Answer:

d. It is equal to the component of the gravitational force acting down the ramp.

Explanation:

The stationary crate is inclined at an angle with the horizontal. The Recall, Frictional Force is any Force that opposes motion.

Because the Force of Friction that is opposing the motion of the crate along the inclination side.

Therefore this Frictional force is balanced or equal to the force that is driving the inclined force.

Hence Frictional Force is equal to the Gravitational Force that is acting in the ramp, that is why the crate is stationery.

8 0
3 years ago
A block lies on a plane raised an angle θ from the horizontal. Three forces act upon the block: F⃗ w, the force of gravity; F⃗ n
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Answer:

YFy = 0 = Ffsinθ + Fncosθ - Fw

Explanation:

From the base of the vector Fn, draw a vertical line. the small angle between this line and Fn is also theta. The component of Fn in the vertical direction is Fncos(theta).

Take a moment to picture extreme cases. Sine is 0 at 0 and 1 at 90. Cosine is 1 at 0 and 0 at 90.

Tilt the incline so that the box is on a flat surface. How much of the gravitational force is along the x direction of the floor.

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How many resistors are found in this circuit? <br> A) 0<br> B) 1<br> C) 2<br> D) 3
Dmitriy789 [7]
Show us the pictures I don't see it
6 0
3 years ago
Across:
vladimir2022 [97]

Answer:

it would be c i just had it

Explanation:

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