Another name for the skin as a whole.

A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carrie

Paul [167]

The range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $$\theta$

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $$\theta$

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $$\theta$

U = Iπ $r^{2}$ B cos $$\theta$

For $$\theta$ = 0° →

U(Max) = MB cos(0) = MB = Iπ $r^{2}$ B = 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

375 π x $10^{-7}$ .

For $$\theta$ = 90° →

U = MB cos (90) = 0

For $$\theta$ = 180° →

U(Min) = MB cos(0) = - MB = - Iπ $r^{2}$ B = - 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

- 375 π x $10^{-7}$ .

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

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3 0

1 year ago

Dr. John Paul Stapp was U.S. Air Force officer whostudied the effects of extreme deceleration on thehuman body. On December 10,

Ira Lisetskai [31]

Answer:

(a) a = 56.4 m/s², his acceleration a, in multiples of gravity g, is 5.76 g

(b) a = -201.43 m/s², his deceleration -a, in multiples of gravity g, is -20.56 g

Explanation:

(a)

When moving upwards, the initial velocity, u = 0 (he accelerated from rest)

When moving upwards, the final velocity, v = 282 m/s

time of motion during this acceleration, t = 5 s

His acceleration is calculated as;

v = u + at

282 = 0 + 5a

a = 282 / 5

a = 56.4 m/s²

Ratio of his acceleration, a to gravity, g = a/g = 56.4 / 9.8 = 5.76

a = 5.76 g

(b)

When moving downwards, the initial velocity, u = 282 m/s

When moving downwards, the final velocity, v = 0 (he was brought to rest)

time of motion during this deceleration, t = 1.4 s

His deceleration is calculated as;

v = u + at

0 = 282 + 1.4a

1.4a = -282

a = -282 / 1.4

a = -201.43 m/s²

Ratio of his deceleration, -a to gravity, g = -a/g = 201.43 / 9.8 = 20.56

a = -20.56 g

3 0

2 years ago

Conductivities are often measured by comparing the resistance of a cell filled with the sample to its resistance when filled wit

Bezzdna [24]

Answer:

1200 Sm^2mol^-1

Explanation:

Given data :

conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1

conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1

Kkcl = 1.1639 - 0.076 = 1.0879 Sm^-1

Resistance = 33.21 Ω

where conductivity can be expressed as = $\frac{Cell constant}{Resistance }$

hence cell constant = conductivity * Resistance

= 1.0879 * 33.21 = 36.13m^-1

conductivity of CH3COOH ( kCH3COOH ) = 36.13 / 300

= 0.120 Sm^-1

<u>Determine the molar conductivity of acetic acid</u>

= ( kCH3COOH * 1000 ) / C

C = 0.1 mol dm

= (0.120 * 1000) / 0.1 = 1200 Sm^2mol^-1

3 0

2 years ago

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out

anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

$0 + mgH = \frac{1}{2}mv^{2} + 0$

$2gH = v^{2}$

$v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s$

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

$mv = (m + m')v'$

$65.0\times 9.89 = (65.0 + 20.0)v'$

$v' = 7.56\ m/s$

Now, time taken for the fall:

$h = \frac{1}{2}gt^{2}$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s$

Now, the horizontal distance is given by:

x = v't = $7.56\times 0.638 = 4.823\ m$

7 0

2 years ago

Read 2 more answers

If the trend changed toward traditional (pre-World War II) families, how would that affect women’s rights?

LUCKY_DIMON [66]

If the trend changed toward traditional (pre-World War II) families, the women’s rights are employment in manufacturing sector.

<h3>What is World war?</h3>

The war between two countries to take over each other's kingdom using weapons to kill each other.

Before the world war, the army needs armor, weapons, guns and tanks. Their manufacturing is only possible with many workers to work for long hours. If the men are not enough, then women are given opportunities to work with them.

Thus, the women’s rights are employment in manufacturing sector when trend changed traditional families.

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7 0

1 year ago