The motion of a piston in an auto engine is simple harmonic. If the piston travels back and forth over a distance of 10 cm and t
1 answer:
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Answer: B. 16
Explanation:
The given table:
x y
2.5 6.25
9.4 88.36
15.6 243.63
19.5 380.25
25.8 665.64
Here, we observed that value of y is the square value of x.
i.e. [For example 2.5²=6.25, 9.4²=88.36, 15.6²=243.63]
Put x=4, we get
Hence, the approximate value of y for x = 4 is 16.
So, the correct option is B. 16 .
Answer:
Ex = kq 2x / ∛ (x² + y²)² and Ex = 2008 N / C
Explanation:
a) The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,
The equation for the electric field produced by a point charge is
E = k q / r²
With r the distance between the point charge and the positive test charge
We look for each electric field
Particle 1. Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance
r² = x² + y²
E1 = k q / (x² + y²)
Particle 2. located at x = -24.9 m
r² = x² + y²
E2 = k q / (x² + y²)
We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.
In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find
Let's measure the angle from axis X
cos θ = CA / H = x / (x2 + y2) ½
E1x = E1 cos θ
E2x = = E1 cos θ
The resulting field
Ey = 0
Ex = E1x + E2x 2 E1x
Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)
Ex = kq 2x / ∛ (x² + y²)²
b) For this part we substitute the numerical values
Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂
Ex = 497.15 x / (x² + 0.062) ³/₂
Point where can the value of the electric field x = 38.1 cm = 0.381 m
Ex = 497.15 0.381 / (0.381² + 0.062) ³/₂
Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2
Ex= 189.41 /0.0943
Ex = 2008 N / C
c) E = 1.00 kN / C = 1000 N / C
To solve this part we must find x in the equation
Ex = 497.15 x / (x² + 0.062) ³/₂
Let's use some arithmetic
Ex / 497.15 = x / (x² + 0.062) ³/₂
[Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃
∛[Ex / 497.15]² = (∛x²) / (x² + 0.062) (1)
The roots of this equation are the solution to the problem,
For Ex = 1.00 kN / C = 1000 N / C
[Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312
1.312 = (∛x² ) / (x² + 0.062)
1.312 (x² + 0.062) = ∛x²
1.312 X² - ∛x² + 1.312 0.062 = 0
1.312 X² - ∛x² + 0.0813 = 0
We need used computer
