Answer: 3.61×10^5 A
Step-by-step explanation: Since the brain has been modeled as a current carrying loop, we use the formulae for the magnetic field on a current carrying loop to get the current on the hemisphere of the brain.
The formulae is given below as
B = u×Ia²/2(x²+a²)^3/2
Where B = strength of magnetic field on the axis of a circular loop = 4.15T
u = permeability of free space = 1.256×10^-6 mkg/s²A²
I = current on loop =?
a = radius of loop.
Radius of loop is gotten as shown... Radius = diameter /2, but diameter = 65mm hence radius = 32.5mm = 32.5×10^-3 m = 3.25×10^-2m
x = distance of the sensor away from center of loop = 2.10 cm = 0.021m
By substituting the parameters into the formulae, we have that
4.15 = 1.256×10^-6 × I × (3.25×10^-2)²/2{(0.021²) + (3.25×10^-2)²}^3/2
4.15 = 13.2665 × 10^-10 × I/ 2( 0.00149725)^3/2
4.15 = 1.32665 ×10^-9 × I / 2( 0.000058)
4.15 × 2( 0.000058) = 1.32665 ×10^-9 × I
I = 4.15 × 2( 0.000058)/ 1.32665 ×10^-9
I = 4.80×10^-4 / 1.32665 ×10^-9
I = 3.61×10^5 A
Answer:
240 feet
Step-by-step explanation:
i multiplied 6
inches and 40 feet.
the answer would be -8<x<2
For this case, the first thing we must do is define variables.
We have then:
x: number of pens
y: number of pencils
We now write the system of inequations:
The solution to the system of inequations is given by the shaded region.
Note: see attached image.
The correct answer is C.
You can tell this by factoring the equation to get the zeros. To start, pull out the greatest common factor.
f(x) = x^4 + x^3 - 2x^2
Since each term has at least x^2, we can factor it out.
f(x) = x^2(x^2 + x - 2)
Now we can factor the inside by looking for factors of the constant, which is 2, that add up to the coefficient of x. 2 and -1 both add up to 1 and multiply to -2. So, we place these two numbers in parenthesis with an x.
f(x) = x^2(x + 2)(x - 1)
Now we can also separate the x^2 into 2 x's.
f(x) = (x)(x)(x + 2)(x - 1)
To find the zeros, we need to set them all equal to 0
x = 0
x = 0
x + 2 = 0
x = -2
x - 1 = 0
x = 1
Since there are two 0's, we know the graph just touches there. Since there are 1 of the other two numbers, we know that it crosses there.