I'm not sure but I had this question on a benchmark I think its the density of the wire you need to find the density or the mass I'm not sure but i do remember this question
Answer:
D) Some of the light passes through, and some of the light is absorbed or scattered by the object.
Explanation:
When light strikes translucent materials, only some of the light passes through them. The light does not pass directly through the materials. ... When light strikes an opaque object none of it passes through. Most of the light is either reflected by the object or absorbed and converted to heat.
(I googled it) ☺
Dependent variable is your answer.
Answer:
<h2>
d₂ = 3d</h2><h2>
The diameter of the second wire is 3 times that of the initial wire.</h2>
Explanation:
Using the formula for calculating the resistivity of an object to find the diameter.
Resistivity P = RA/L
R is the resistance of the material
A is the cross sectional area
L is the length of the material
Since A = πd²/4
P = R( πd²/4)/L
P = Rπd²/4L ... 1
If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;
P₂ = (R/9)A₂/L₂
P₂ = (R/9)(πd₂²/4)/L₂
P₂ = (Rπd₂²/36)/L₂
P₂ = (Rπd₂²)/36L₂
Since the length and resistivity are the same;
P = P₂ and L =L₂
Equating 1 and 2;
Rπd²/4L = (Rπd₂²)/36L₂
Rπd²/4L = (Rπd₂²)/36L
d² = d₂²/9
d₂² = 9d²
Taking the square root of both sides;
√d₂² = √9d²
d₂ = 3d
Therefore the diameter of the second wire is 3 times that of the initial wire
Answer:
protons
atomic number
Note: The same atomic number can be associated with several different values of atomic mass, but an element can have only 1 atomic number,
