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2 years ago

10

The magnetic moment of a rectangular loop that carries 6 A and has dimensions are 0.04 m × 0.08 m is 12 Am2. How many turns are

present?

1 answer:

Paul [167]2 years ago

5 0

Answer:

The number of turns present in the loop are 625 turns.

Explanation:

Given;

current on the rectangular loop, I = 6A

Dimension of the rectangular loop, L x B = 0.04 m × 0.08 m

Magnetic moment of the loop, Pm = 12 Am²

Area of the rectangular loop, A = L x B = 0.04 m × 0.08 m = 0.0032 m²

The magnetic moment of plane carrying coil/loop is given as;

Pm = NIA

Where;

N is the number of turns

A is the area of the loop

I is the current on the loop

$N = \frac{P_m}{IA} \\\\N = \frac{12}{6*0.0032}\\\\N = 625 \ turns$

Therefore, the number of turns present in the loop are 625 turns.

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Two ice skaters, Lilly and John, face each other while stationary and push against each others hands. John's mass is twice the m

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Answer:

lily's speed would be twice john's speed

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2 years ago

A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float

storchak [24]

Answer:

t ’= $\frac{1450}{0.6499 + 2 v_r}$ , v_r = 1 m/s t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

v_b - v_r = d / t

By the time the boat goes down the river

v_b + v_r = d '/ t'

let's subtract the equations

2 v_r = d ’/ t’ - d / t

d ’/ t’ = 2v_r + d / t

$t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }$

In the exercise they tell us

d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

t ’= $\frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \ + 2 \ v_r}$

t ’= $\frac{1450}{0.6499 + 2 v_r}$

In order to complete the calculation, we must assume a river speed

v_r = 1 m / s

let's calculate

t ’= $\frac{ 1450}{ 0.6499 + 2 \ 1}$

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8 0

2 years ago

The Americium nucleus, 241 95 Am, decays to a Neptunium nucleus, 237 93 Np, by emitting an alpha particle of mass 4.00260 u and

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Answer:

The mass of Neptunium is 237.054 u.

Explanation:

Given that,

Mass of Americium = 241.05682 u

Mass of alpha particle = 4.00260 u

The equation is,

$_{95}^{241}Am\rightarrow _{93}^{237}NP+\alpha\ particle+5.5 MeV$

Let the mass of Neptunium is m.

Since the mass remain same.

We need to calculate the mass of Neptunium

Using formula of mass

Mass of Neptunium = Mass of Americium -Mass of alpha particle

Put the value into the formula

$m=241.05682 -4.00260$

$m=237.054\ u$

Hence, The mass of Neptunium is 237.054 u.

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2 years ago

Use what you have learned in the reading to answer the following question. The point of a compass needle points to Earth’s north

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1 year ago

The perihelion of the comet TOTAS is 1.69 AU and the aphelion is 4.40 AU. Given that its speed at perihelion is 28 km/s, what is

dybincka [34]

Answer:

The speed at the aphelion is 10.75 km/s.

Explanation:

The angular momentum is defined as:

$L = mrv$ (1)

Since there is no torque acting on the system, it can be expressed in the following way:

$t = \frac{\Delta L}{\Delta t}$

$t \Delta t = \Delta L$

$\Delta L = 0$

$L_{a} - L_{p} = 0$

$L_{a} = L_{p}$ (2)

Replacing equation 1 in equation 2 it is gotten:

$mr_{a}v_{a} =mr_{p}v_{p}$ (3)

Where m is the mass of the comet, $r_{a}$ is the orbital radius at the aphelion, $v_{a}$ is the speed at the aphelion, $r_{p}$ is the orbital radius at the perihelion and $v_{p}$ is the speed at the perihelion.

From equation 3 v_{a} will be isolated:

$v_{a} = \frac{mr_{p}v_{p}}{mr_{a}}$

$v_{a} = \frac{r_{p}v_{p}}{r_{a}}$ (4)

Before replacing all the values in equation 4 it is necessary to express the orbital radius for the perihelion and the aphelion from AU (astronomical units) to meters, and then from meters to kilometers:

$r_{p} = 1.69 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $2.528x10^{11} m$

$r_{p} = 2.528x10^{11} m x \frac{1km}{1000m}$ ⇒ $252800000 km$

$r_{a} = 4.40 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $6.582x10^{11} m$

$r_{p} = 6.582x10^{11} m x \frac{1km}{1000m}$ ⇒ $658200000 km$

Then, finally equation 4 can be used:

$v_{a} = \frac{(252800000 km)(28 km/s)}{(658200000 km)}$

$v_{a} = 10.75 km/s$

Hence, the speed at the aphelion is 10.75 km/s.

8 0

2 years ago