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Answer:
So after stretching new resistance will be 0.1823 ohm
Explanation:
We have given initially length of the wire
Radius of the wire
Resistance of the wire initially
We know that resistance is equal to ,here
is resistivity, l is length and A is area
From the relation we can say that
Now length of wire become 6.75 m
Volume will be constant
So
So
So
The focal length of a lens needed by a woman whose near point is 50cm from her eyes is 50cm.
To find the answer, we have to know about the focal length of correcting lens.
<h3>How to find the focal length of correcting lens?</h3>- If x is the distance of nearest point of the defective eye and D is the least distance of distinct vision, then, the expression for focal length of the correcting lens will be,
- It is given that, the woman whose near point is 50cm from her eyes, assuming the least distance of distinct vision for a normal eye is 25cm. Thus, the focal length will be,
Thus, we can conclude that, the focal length of a lens needed by a woman whose near point is 50cm from her eyes is 50cm.
Learn more about the focal length here:
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Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C