A 2100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact w
1 answer:
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Answer:
A. Fnx = 5.71*10⁻⁵ N , Fny= -3.67*10⁻⁵ N
B. Fn= 6.78 *10⁻⁵ N
C. α= 32.4° counterclockwise with the positive x+ axis
Explanation:
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
* 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
*10⁻⁵ N
Fn= 6.78 *10⁻⁵ N
Fn direction (α)
α= -32.4°
α= 32.4° counterclockwise with the positive x+ axis
To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.
Planet gravitational force
Distance between planet and star
Gravitational force is
Applying the new distance,
Replacing with the previous force,
Replacing our values
Therefore the magnitude of the force on the star due to the planet is