A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on
1 answer:
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Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)
q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
Answer:
Same frequency, shorter wavelength
Explanation:
The speed of a wave is given by
where,
f = Frequency
= Wavelength
It can be seen that the wavelength is directly proportional to the velocity.
Here the frequency of the sound does not change.
But the velocity of the sound in air is slower.
Hence, the frequency remains same and the wavelength shortens.
Answer:
L = 0.635m
Explanation:
This problem involves the concept of stationary waves in pipes. For pipes closed at one end,
The frequency f = nv/4L for n = 1,3,5....n
For pipes open at both ends
f = nv/2L for n = 1,2,3,4...n
Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.
The film solution can be found in the attachment below.
Answer:
the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)
the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)
Explanation:
Use Biot, Savart, the magnetic field
Given that,
i = 1.00A
d → l = 4.00 m m ^ j
r = 2.5m
Displacement vector is
=2.5m
on the axis of x at x = 2.5
r = 2.5m
And unit vector
Therefore, the magnetic field is as follow
(Along z direction)
B)r = 5.00m
Displacement vector is
=5.00m
on the axis of x at x = 5.0
r = 5.00m
And unit vector
Therefore, the magnetic field is as follow
(Along x direction)