2.

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

$d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }$

$d_{13} =\sqrt{24.68}$ * 10⁻²m = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² = ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) = - 3.67*10⁻⁵ N

F₂₃x = F₂₃ = +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x = - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

$F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2} }$

$F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2} }$ *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction (α)

$\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )$

$\alpha =tan^{-1}( \frac{-3.67 }{5.71} )$

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0

3 years ago

A space walking astronaut has become detached from her spaceship.shes floating in space while holding a wrench she is thinking a

victus00 [196]

Answer:

Newton's third law of motion states that for every action, there is equal and opposite reaction.

While space walking, when the astronaut gets detached from the space ship, she floats in space holding a wrench. In order to get back to the spaceship, she should throw the wrench in the opposite direction of the spaceship. This action would cause a reaction on her own body and she would be pushed away from the wrench and towards the spaceship. Thus, she can return back to the spaceship in this way.

4 0

3 years ago

In medieval warfare, one of the greatest technological advancement was the trebuchet. The trebuchet was used to sling rocks into

NNADVOKAT [17]

Answer:

2) a_y= -g 3) vₓ=constant v_y = v_{oy} - g t, 4) vₓ = v₀ₓ - ax t

5) changes the horizontal speed, should change range

7) changes the vertical speed change the maximum height

Explanation:

1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.

2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration

3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression

v_y = v_{oy} - gt

at the point of maximum height, vy = 0 is equal to the maximum height

4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed

In the graph it would be directed to the left, therefore the velocity would be

vₓ = v₀ₓ - ax t

5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.

the equations of motion are

x = v₀ₓ t

y = v_{oy} t - ½ g t²

7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,

the equations of motion are the same.

4 0

3 years ago