Answer:
l= 4 mi : width of the park
w= 1 mi : length of the park
Explanation:
Formula to find the area of the rectangle:
A= w*l Formula(1)
Where,
A is the area of the rectangle in mi²
w is the width of the rectangle in mi
l is the width of the rectangle in mi
Known data
A = 4 mi²
l = (w+3)mi Equation (1)
Problem development
We replace the data in the formula (1)
A= w*l
4 = w* (w+3)
4= w²+3w
w²+3w-4= 0
We factor the equation:
We look for two numbers whose sum is 3 and whose multiplication is -4
(w-1)(w+4) = 0 Equation (2)
The values of w for which the equation (2) is zero are:
w = 1 and w = -4
We take the positive value w = 1 because w is a dimension and cannot be negative.
w = 1 mi :width of the park
We replace w = 1 mi in the equation (1) to calculate the length of the park:
l= (w+3) mi
l= ( 1+3) mi
l= 4 mi
Answer:
Sounds cool.. but what do they do?
Explanation:
Answer:
V = 9.682 × 10^(-6) V
Explanation:
Given data
thick = 190 µm
wide = 4.20 mm
magnitude B = 0.78 T
current i = 32 A
to find out
Calculate V
solution
we know v formula that is
V = magnitude× current / (no of charge carriers ×thickness × e
here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³
so put all value we get
V = magnitude× current / (no of charge carriers ×thickness × e
V = 0.78 × 32 / (8.47 x 10^28 × 190 × 1.602 x 10^(-19)
V = 9.682 × 10^(-6) V
Answer:
I can't see the attachment it's too blury I can see bout two words and thats it
Explanation:
Think i got an eye infection or bad eye sight
Answer:
a. the bottom medium
Explanation:
it has the least index of refraction and hence most rarer.
