How do you find distance on a velocity time graph?
1 answer:
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Answer:
R = 25.3ohms
Explanation:
See attachment below.
Answer:
3.75 × 10⁻⁸ N
Explanation:
Given:
Intensity of the electromagnetic wave, I = 150 W/m²
Sides of the board = 25 cm (= 0.25 m) and 30 cm (= 0.30 m)
therefore,
the area of the rectangular box, A = 0.25 × 0.30 = 0.075 m²
Now,
force exerted on the card by the radiation, F =
here,
C is the speed of the light = 3 × 10⁸ m/s
on substituting the respective values, we get
F =
or
F = 3.75 × 10⁻⁸ N
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.
