Answer:
131.2 kg, to have the same precision as 58.2 kg
Explanation:
A charge of +80 µC is placed on the x axis at x = 0. A second charge of –50 µC is placed on the x axis at x = 50 cm. What is the magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x axis at x = 30 cm?
Answer:
Explanation:
77
The magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x-axis at x = 30 cm is equal to 77 N.
What is coulomb's law?According to Coulomb’s law, the force of repulsion or attraction between two charged bodies is the multiplication of their charges and is inversely proportional to the square of the distance between them.
The magnitude of the electric force can be written as follows:
[tex]F =k \frac{q_1q_2}{r^2}[/tex]
where k has a value of 9 × 10⁹ N.m²/C².
Given, the first charge, q₁ = + 80 ×10⁻⁶ C
The second charge , q₂ = - 50 × 10⁻⁶ C
The third charge, q₃ = + 4 ×10⁻⁶ C
The distance between these two charges (q₁ and q₃) , r = 0.30 m
The distance between these two charges (q₂ and q₃) , r = 0.20m
The magnitude of the electrostatic force on the third charges will be:
[tex]F =9\times 10^9 (\frac{80\times 10^{-6}C\times 4 \times 10^{-6}C}{(0.30)^2} ) +9\times 10^9 (\frac{50\times 10^{-6}C\times 4 \times 10^{-6}C}{(0.20)^2} )[/tex]
F = 77 N
Therefore, the magnitude of the electrostatic force is equal to 77 N.
Learn more about Coulomb's law, here:
brainly.com/question/506926
#SPJ2
A block of mass m = 2.5 kg is attached to a spring with spring constant k = 730 N/m. It is initially at rest on an inclined plane that is at an angle of θ = 21° with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is μk = 0.19. In the initial position, where the spring is compressed by a distance of d = 0.11 m, the mass is at its lowest position and the spring is compressed the maximum amount. Take the initial gravitational energy of the block as zero.
A) What is the block's initial mechanical energy?
B) If the spring pushes the block up the incline, what distance, L, in meters will the block travel before coming to rest? The spring remains attached to both the block and the fixed wall throughout its motion.
Answer:
A) Em = 4.41 J
B) L = 0.33m
Explanation:
A) The total mechanical energy of the block is the elastic potential energy due to the compressed spring. The gravitational energy is zero. Then you have:
[tex]E_m=\frac{1}{2}k(\Delta x)^2[/tex]
k: constant's spring = 730 N/m
Δx: distance of the compression = 0.11m
You replace the values of k and Δx:
[tex]E_m=\frac{1}{2}(730N/m)(0.11m)^2=4.41\ J[/tex]
B) To find the distance L traveled by the block you take into account that the total mechanical energy of the block is countered by the work done by the friction force, and also by the work done by the gravitational energy.
Then, you have:
[tex]E_m-W_f-W_g=0\\\\W_f=(\mu Mg cos\theta)L\\\\W_g=(Mgsin\theta)L[/tex]
μ: coefficient of kinetic friction = 0.19
g: gravitational acceleration = 9.8m/s^2
M: mass of the block = 2.5kg
θ: angle of the inclined plane = 21°
You replace the values of all parameters:
[tex]E_m-W_f-W_g=0\\\\4.41-(0.19)(2.5kg)(9.8m/s^2)(cos21\°)L-(2.5kg)(9.8m/s^2)(sin21\°)L=0\\\\4.41-4.34L-8.78L=0\\\\4.41-13.12L=0\\\\L=0.33m[/tex]
hence, the distance L in which the block stops is 0.33m
(a) The block's initial mechanical energy on the given position is 4.42 J.
(b) The distance traveled by the block when it is pushed by the spring before coming to rest is 1.02 m.
The given parameters;
mass of the block, m = 2.5 kgspring constant, k = 730 N/mangle of inclination, θ = 21°coefficient of friction, μ = 0.19compression of the spring, x = 0.11 mThe block's initial mechanical energy is calculated as follows;
[tex]E = K.E _i + P.E_i\\\\E = \frac{1}{2} mv^2 \ + \ \frac{1}{2} kx^2\\\\E = \frac{1}{2} m (0)^2 \ + \ \frac{1}{2} \times 730 \times (0.11)^2\\\\E = 4.42 \ J[/tex]
The block will travel up if the energy applied by the spring is greater than the work-done by frictional force on the block.
The work-done on the block by the frictional force is calculated as follows;
[tex]W_f = F_k \times d\\\\W_f= \mu_k F_n \times d\\\\W_f = \mu_k mgcos(\theta) \times d\\\\W_f = (0.19)(2.5)(9.8)cos(21) \times d \\\\W_f = 4.346 d[/tex]
Apply work-energy theorem;
[tex]4.346d = 4.42\\\\d = \frac{4.42}{4.346} = 1.02 \ m[/tex]
Thus, the distance traveled by the block when it is pushed by the spring before coming to rest is 1.02 m.
Learn more here:https://brainly.com/question/15583298