Answer:
Diagrammatic representation
'The normal is a line perpendicular to the surface of the mirror'.This is the correct statement that corrects an error on the site.
<h3>What is the law of reflection?</h3>
The law of reflection specifies that upon reflection from a downy surface, the slope of the reflected ray is similar to the slope of the incident ray.
The reflected ray is consistently in the plane determined by the incident ray and perpendicular to the surface at the point of reference of the incident ray.
When the light rays descend on the smooth surface, the angle of reflection is similar to the angle of incidence, also the incident ray, the reflected ray, and the normal to the surface all lie in a similar plane.
Hence 'The normal is a line perpendicular to the surface of the mirror'.This is the correct statement that corrects an error on the site
To learn more about the law of reflection refer to the link;
brainly.com/question/12029226
Answer:
Neon (Ne), chemical element, inert gas of Group 18 (noble gases) of the periodic table, used in electric signs and fluorescent lamps. Colourless, odourless, tasteless, and lighter than air, neon gas occurs in minute quantities in Earth's atmosphere and trapped within the rocks of Earth's crust.
Explanation:
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Answer: I think its 120
Explanation: thx for the free points :)
Explanation:
Given that,
Terminal voltage = 3.200 V
Internal resistance 
(a). We need to calculate the current
Using rule of loop


Where, E = emf
R = resistance
r = internal resistance
Put the value into the formula


(b). We need to calculate the terminal voltage
Using formula of terminal voltage

Where, V = terminal voltage
I = current
r = internal resistance
Put the value into the formula


(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf


Hence, This is the required solution.