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Nookie1986 [14]
3 years ago
11

A long wire is known to have a radius greater than 4.0 mm and to carry a current uniformly distributed over its cross section. i

f the magnitude of the magnetic field is 0.285 mt at a point 4.0 mm from the axis of the wire and 0.200 mt at a point 10 mm from the axis, what is the radius of the wire?
Physics
1 answer:
ollegr [7]3 years ago
8 0

Magnetic field outside it due to long wire is given by

B = \frac{u_o i}{2 \pi r}

Magnetic field due to long wire inside wire at any point

B = \frac{u_o i r}{2 \pi R^2}

Now the ratio of two magnetic field is given by

\frac{B_{in}}{B_{out}} = \frac{r_1/R^2}{1/r_2}

\frac{0.285}{0.200} = \frac{4*10}{R^2}

1.425 = \frac{40}{R^2}

R = 5.3 mm

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A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons
Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

H_m=1.65m

H_E=1.16307m

Explanation:

From the question we are told that

Mass of ball M=2kg

Length of string L= 2m

Wind force F=13.2N

Generally the equation for \angle \theta is mathematically given as

tan\theta=\frac{F}{mg}

\theta=tan^-^1\frac{F}{mg}

\theta=tan^-^1\frac{13.2}{2*2}

\theta=73.14\textdegree

Max angle =2*\theta= 2*73.14=>146.28\textdegree

Generally the equation for max Height H_m is mathematically given as

H_m=L(1-cos146.28)

H_m=0.9(1+0.8318)

H_m=1.65m

Generally the equation for Equilibrium Height H_E is mathematically given as

H_E=L(1-cos73.14)

H_E=0.9(1+0.2923)

H_E=1.16307m

8 0
2 years ago
Consider a single-slit diffraction pattern for λ=589nm , projected on a screen that is 1.00 m from a slit of width 0.25 mm. How
const2013 [10]

Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm

Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as

y = R×(mλ/d)

Where y = distance between nth fringes and the center fringe.

m = order of fringe

λ = wavelength of light = 589nm = 589×10^-9m

R = distance between slits and screen = 1.0m

d = distance between slits = 0.25mm = 0.00025m

For distance between the first dark fringe and the center fringe.

This implies that m = 1

y = 1 × 589×10^-9 × 1/0.00025

y = 589×10^-9/0.00025

y = 2,356,000 × 10^-9

y = 2.356 × 10^-3 m = 2.356mm

For the second dark fringe, this implies that m = 2

y = 1 × 2 × 589×10^-9/0.00025

y = 1178 × 10^-9 /0.00025

y = 4,712,000 × 10^-9

y = 4.712 × 10^-3 m = 4.712mm

8 0
3 years ago
Find the GCF 12, 18, and 84
ahrayia [7]
To find the answer, plot down the factors for every number.

12:   1, 2 ,3 ,4, 6, 12

18:   1, 2, 3, 6, 9, 18

84:  1, 2, 3, 4, 6, 7, 12

If you noticed, the number that was common to the 3 numbers, were 1, 2, 3, and 6
And 6 is the bigger number
So 6 is your GCF
5 0
3 years ago
Read 2 more answers
Which of the following examples illustrates static friction in action?
lilavasa [31]
I think its d. but im not sure
3 0
3 years ago
Read 2 more answers
Pls answer it plssss
Sonbull [250]

Answer:

1.Stronger bones 2.Joint flexibility

3 0
2 years ago
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