Answer this for me please and thank you

A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the

SOVA2 [1]

Answer:

Johnny created an electromagnet out of a solenoid (a coil of wires with 20 loops), an iron core (of 1 nail), and a single 9 V battery. When Johnny does this, he creates a small magnetic field that allows him to pick up 2 paper clips. Using a CER format, explain to Johnny three things he could change that would increase the strength of his magnetic field and why each change increases the magnetic field. You may want to write three paragraphs to make this easier for the reader to understand

5 0

3 years ago

What is the mass of a 735-N weight on Earth?

Allisa [31]

W = mg, Assuming g ≈ 9.8 m/s² on the earth surface.

735 N = m* 9.8

735/9.8 = m

75 = m

Mass , m = 75 kg. B.

3 0

3 years ago

Juans mother drives 7.25 miles southwest to her favorite shopping mall. What is the average velocity of her automobile if she ar

Irina-Kira [14]

21.75 Miles Per Hour

I got this by multiplying 7.25(3) because I know 20 minutes is 1/3 of 1 he

5 0

3 years ago

Why do different substances have different melting and boiling points?

azamat

Because the atoms and molecules all have different properties

3 0

3 years ago

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

a

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

b

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$