how does the frequency of a radio wave compare to the frequency of the vibrating electrons that produce it?
1 answer:
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Answer:
a = 0.77 m/s^2
Explanation:
The centripetal acceleration is given as
The velocity of the car can be calculated using the circumference of the track and the period of the car.
So, the acceleration is
Answer:
1) E = 2.25 i^+ 0.809j^) 10⁹ N / C , 2) E = 2.39 10⁹ N / C , 3) θ = 19.8º , 4) F = 19.12 10⁹ N , 5) E = (1.32 i^+ 3.56 j^) 109 N/C
Explanation:
1) The equation for the electric field is
E = k q / r²
Where K is the Coulomb constant that is worth 8.99 10⁹ N m² /C², q is the load and r is the distance of the load to the test point
Since the electric field is a vector magnitude, we can find its component
X axis
Ex = k q / x²
where the distance on the axis is
x = √ (X₂-x₁)²
x = √ (-15 + 9)² = 6 m
Eₓ = 8.99 10⁹ 9/6²
Eₓ = 2.25 10⁹ N /C
Y Axis
y = √ (y₂-y₁)² = √ (15-5)² = 10 m
= 8.99 10⁹ 9/10²
= 0.809 10⁹ N / C
E = Eₓ i^+ j^
E = 2.25 i^+ 0.809j^) 10⁹ N / C
2) the magnitude can be found using the Pythagorean triangle
E = √ (Eₓ² + ²)
E = √ (2.25² + 0.809²) 10⁹
E = 2.39 10⁹ N / C
3) to find the angle let's use trigonometry
tan θ = / Eₓ
θ = tan⁻¹ / Eₓ
θ = tan⁻¹ (0.809 / 2.25)
θ = 19.8º
Regarding the positive side of the x axis
4) a charge q2 = 8C is placed, let's calculate the force
F = q E
F = 8 2.39 10⁹
F = 19.12 10⁹ N
5) The total electric field at the origin, let's look for its components
q₁ = 9C
r₁ = -9 i ^ + 5 j ^
q₂ = 8 C
r₂ = -15 i ^ + 15 j ^
X axis
Eₓ = E₁ₓ + E₂ₓ
Eₓ = k q₁ / Dx₁² + k q₂ / Dx₂²
Eₓ = 8.99 10⁹ (9 / (9-0)² + 8 / (15-0)²)
Eₓ = 1.32 109 N / A
Y Axis
=
+
= k q₁ / Δy₁² + k q₂ / Δy₂²
= 8.99 109 (9/5² + 8/15²)
= 3.56 109 N / A
E = (1.32 i^+ 3.56 j^) 109 N/C