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kupik [55]
2 years ago
12

Consider two parallel wires where the magnitude of the left currentis 2 I0(io) and that of the right current is I0(io). Point A

is midway between the wires,and B is an equal distance on the other side of the wires.
The ratio ofthe magnitude of the magnetic field at point A to that at point Bis________
Physics
1 answer:
balu736 [363]2 years ago
3 0

Answer:

Explanation:

At the point midway between wires

magnetic field due to wire having current 2I₀

= 10⁻⁷ x 2 x2I₀ / r     where 2r is the distance between wires .

magnetic field due to wire having current I₀

= 10⁻⁷ x 4 I₀ / r

magnetic field due to wire having current I₀

= 10⁻⁷ x 2I₀ / r    

= 10⁻⁷ x 2 I₀ / r     where 2r is the distance between wires .

these fields are in opposite direction as direction of current is same in both .

net magnetic field = (4 - 2 )x 10⁻⁷ x I₀ / r

= 2 x 10⁻⁷ x  I₀ / r

At point A net magnetic field = 2 x 10⁻⁷ x  I₀ / r

At point B , we shall calculate magnetic field

magnetic field due to nearer wire having current  2 I₀ = 10⁻⁷ x 4 I₀ / r

magnetic field due to wire far away = 10⁻⁷ x 2 I₀ / 3r

These magnetic fields act in the same direction so they will add up

net magnetic field = [ (4 I₀ / r)  + (2 I₀ / 3r) ] x 10⁻⁷

= (14 I₀ / 3r ) x 10⁻⁷

Magnetic field at point B = (14 I₀ / 3r ) x 10⁻⁷

Ratio of field at A and B

= 3 / 7 . Ans

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lukranit [14]

Answer:

By 16.7% or 0.167 IPM

Explanation:

Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.

4 0
2 years ago
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3. You have three stars. Star A has an apparent magnitude of 7, Star B has an apparent magnitude of 2, and Star C has an apparen
Mrrafil [7]

"Apparent magnitude" means how bright a star looks to
a person on Earth.

-- The star that appears brightest is the one with the
lowest-number apparent magnitude . . . Star-C, at -4 .

-- All of them are visible from Earth, but may require some 'help'. 
The dimmest stars visible with good human eyes under dark,
non-polluted skies are those with apparent magnitude around 6.
Stars B and C would be visible to the unaided eye, but Star-A
would require binoculars.
Around here, a few miles outside of the Chicago city limits, we're
lucky to see Magnitude-4 without binoculars.

-- It's not possible to determine which star has the highest luminosity.
The apparent magnitude depends on the star's distance from Earth
as well as its luminosity.
A flashlight 3 feet from your face appears much brighter than any
star, although any star is more luminous than the flashlight. 
Distance from you has a lot to do with it.
_____________________________________________

"Absolute magnitude" means how bright each star would appear
to a person on Earth if all stars were at the same distance from us.
(The distance happens to be 32.6 light years.)  It only depends on
the star's real luminosity, not on its distance.

-- It's not possible to determine which star appears brightest.
Star-C (absolute -7) would appear brightest if all stars were
equal distances from us.  But a flashlight ... which has a huge-
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6 0
2 years ago
The free-body diagram of a crate is shown. What is the net force acting on the crate? 352 N to the left 176 N to the left 528 N
Umnica [9.8K]

As per given conditions there are two directions along which forces are acting

1. Net force along left direction is given as

F_{left} = 352N + 176 N = 528 N

2. Net force towards right direction is given as

F_{right} = 528N + 440 N = 968 N

now since the two forces here in opposite direction so here we will have net force given as

F_{net} = F_{right} - F_{left}

F_{net} = 968 - 528

F_{net} = 440 N

so here net forces must be 440 N towards right

7 0
2 years ago
Read 2 more answers
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
2 years ago
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jenyasd209 [6]
What team are you talking about
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2 years ago
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