When drawing a free body diagram, which of the following things would NOT be included

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0

3 years ago

The local high school is installing new bleachers at the stadium and must also add handrails to meet code. The students know the

Advocard [28]

Answer:

The handrails must be approximately 10.63 meters long

Explanation:

The given parameters are;

The height of the bleachers, h = 8 m

The depth of the bleachers, d = 7 m

The length of the hand rails to go along the bleachers from bottom to top is given by Pythagoras' Theorem as follows;

The length of the hand rail = √(d² + h²)

∴ The length of the hand rail = √(7² + 8²) = √113 ≈ 10.63

In order for the handrails to go along the bleachers from top to bottom, they must be approximately 10.63 meters long.

5 0

3 years ago

A cube of side 0.2 m rests on the floor as shown.

andrey2020 [161]

Pressure= hqg
H=depth
q=density
g=gravity

h=0.2
q=7
g=10

0.2*7*10= 14pa

FINAL ANSWER = 14pa

4 0

1 year ago

An engineer is investigating potential energy with two

miv72 [106K]

Answer:

Since they move in the same direction, share the same magnetic qualities, and move one unit identically
We can safely say its D

4 0

3 years ago

Read 2 more answers

Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is λA = 622nm. The other wavelength is λ

Sergio039 [100]

Answer:

$\lambda_{B}=414.67 nm$

Explanation:

In this question we have given

$\lambda_{A}=622nm$

we have to find

$\lambda_{B}=?$

We know that

optical path difference for bright fringe is given as $=n\lambda$

Here,

n is order of fringe

and optical path difference for dark fringe is given as $=(n+.5)\lambda$

since the light with wavelength $\lambda_{A}$ produces its third-order bright fringe at the same place where the light with wavelength $\lambda_{B}$ produces its fourth dark fringe

it means

optical path difference for 3rd order bright fringe= optical path difference for forth order dark fringe

Therefore,

$3\lambda_{A}=(4+.5)\lambda_{B}$ ...............(1)

Put value of $\lambda_{A}$ in equation (1)

$3 \times 622=(4+.5)\lambda_{B}$

$1866=4.5\lambda_{B}$

$\lambda_{B}=414.67 nm$

3 0

3 years ago