A sailboat runs before the wind with a constant speed of 2.8 m/s in a direction 52° north of west. How far (a) west and (b) nort
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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
Answer:
A. A = 0.913 m
B. amax = 132.24m/s^2
C. Fmax = 324.01N
Explanation:
When the block is moving at the equilibrium point , its velocity is maximum.
A. To find the amplitude of the motion you use the following formula for the maximum velocity:
(1)
vmax = maximum velocity = 11.0 m/s
A: amplitude of the motion = ?
w: angular frequency = ?
Then, you have to calculate the angular frequency of the motion, by using the following formula:
(2)
k: spring constant = 355 N/m
m: mass of the object = 2.54 kg
Next, you solve the equation (1) for A and replace the values of vmax and w:
The amplitude of the motion is 0.913m
B. The maximum acceleration of the block is given by:
The maximum acceleration is 132.24 m/s^2
C. The maximum force is calculated by using the second Newton law and the maximum acceleration:
It is also possible to calculate the maximum force by using:
Fmax = k*A = (355N/m)(0.913m) = 324.01N
The maximum force exertedbu the spring on the object is 324.01 N