Answer:
A: They produce a real image.
Explanation:
The images formed on the retina of the eye for a normal visibility must always be real.
Only a real image can be physically projected on any physical object whereas the virtual images are visible due to reflections.
- The nearsightedness is corrected with the help of a concave lens since it is the condition of the eye lens remaining thick and curved to converge the rays entering the eyes after a shorter distance which results in their image formation even before the retinal surface so to initially diverge them a bit so that they converge on the retinal surface and form the image there we use concave lens. Vice-versa of the above justification in the case of farsightedness.
D = 497.4x10⁻⁶m. The diameter of a mile of 24-gauge copper wire with resistance of 0.14 kΩ and resistivity of copper 1.7×10−8Ω⋅m is 497.4x10⁻⁶m.
In order to solve this problem we have to use the equation that relates resistance and resistivity:
R = ρL/A
Where ρ is the resistivity of the matter, the length of the wire, and A the area of the cross section of the wire.
If a mile of 24-gauge copper wire has a resistance of 0.14 kΩ and the resistivity of copper is 1.7×10⁻⁸ Ω⋅m. Determine the diameter of the wire.
First, we have to clear A from the equation R = ρL/A:
A = ρL/R
Substituting the values
A = [(1.7×10⁻⁸Ω⋅m)(1.6x10³m)]/(0.14x10³Ω)
A = 1.9x10⁻⁷m²
The area of a circle is given by A = πr² = π(D/2)² = πD²/4, to calculate the diameter D we have to clear D from the equation:
D = √4A/π
Substituting the value of A:
D = √4(1.9x10⁻⁷m²)/π
D = 497.4x10⁻⁶m
<span>Heat from the Sun is transferred to the sand without direct contact. This heat is then transferred to your feet by direct contact.</span>
The angle is approximately 81 degrees. Since the person's eyes are 4 feet above the ground level, and the top of the building is 190 feet tall, the vertical distance between the person's eyes and the top of the building is 190-4=186 feet. The horizontal distance is 30 feet, since the person is standing at the end of the shadow. Therefore tan(angle)=186/30, the angle is equal to arctan(186/30)=81, approximately.