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3 years ago

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A solenoid has 332 turns and a length of 14 cm. If a current of 0.88 A produces a magnetic flux density of 0.28 T in the core of

the solenoid, what is the relative permeability of the core material?

1 answer:

irina1246 [14]3 years ago

3 0

Answer:

106.83

Explanation:

N = 332, l = 14 cm = 0.14 m, i = 0.88 A, B = 0.28 T

Let ur be the relative permeability

B = u0 x ur x n x i

0.28 = 4 x 3.14 x 10^-7 x ur x 332 x 0.88 / 0.14 ( n = N / l)

ur = 106.83

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Need some help with these two physics problems!

Juliette [100K]

The force that keeps the puck moving is 0.25 N while the velocity of the puck is 3.7 m/s.

<h3>What is the centripetal force?</h3>

We know that the centripetal force is the force that acts on a body that is moving along a circular path. In this case, we are told that the puck is moving along a circular path hence it is acted upon by the centripetal force that acts on it.

The centripetal force in this case would be supplied by the weight of the object that is moving in the circular path. Thus we can write in our equation that;

Centripetal force = Weight of object = mg

m = mass of the object

g = acceleration due to gravity

Then;

W = 0.026 Kg * 9.8 m/s^2

W = 0.25 N

To obtain the velocity of the object;

FT = mv^2/r

v = √ FT r/m

v = √0.25 * 1.4/0.026

v = 3.7 m/s

Learn more about centripetal force:brainly.com/question/11324711

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1 year ago

At t= 0, a particle moving in the xy plane with constant acceleration has a velocity of Vi= (3.00i -2.00j) m/s and is at the ori

ivanzaharov [21]

Answer:

(a) a = (2i + 4.5j) m/s^2

(b) r = ro + vot + (1/2)at^2

Explanation:

(a) The acceleration of the particle is given by:

$\vec{a}=\frac{\vec{v}-\vec{v_o}}{t}\\\\$

vo: initial velocity = (3.00i -2.00j) m/s

v: final velocity = (9.00i + 7.00j) m/s

t = 3s

by replacing the values of the vectors and time you obtain:

$\vec{a}=\frac{1}{3s}[(9.00-3.00)\hat{i}+(7.00-(-2.00))\hat{j}]\\\\\vec{a}=(2\hat{i}+4.5\hat{j})m/s^2$

(b) The position vector is given by:

$\vec{r}=\vec{r_o}+\vec{v_o}t+\frac{1}{2}\vec{a}t^2$

where vo = (3.00i -2.00j) m/s and a = (2.00i + 4.50j)m/s^2

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2 years ago

Two particles are fixed to an x axis: particle 1 of charge −8.00 ✕ 10⁻⁷ C at x = 6.00 cm, and particle 2 of charge +8.00 ✕ 10⁻⁷

victus00 [196]

Answer:

0 N/C

Explanation:

Parameters given:

$q_1 = -8.00 * 10^{-7} C$

$x_1 = 6.00 cm$

$q_2 = +8.00 * 10^{-7} C$

$x_2 = 21 cm$

The distance between $q_1$ and $q_2$ is

$21 - 6 = 15cm$

Electric field is given as

$E = \frac{kq}{r^2}$

r = 15/2 = 7.5cm = 0.075m

The electric field at their midpoint due to $q_1$ is:

$E = \frac{9 * 10^9 * -8.0 * 10^{-7}}{0.075^2}$

$E_1$ = $-1.28 * 10^6 N/C$

The electric field at the midpoint due to $q_2$ is:

$E = \frac{9 * 10^9 * 8.0 * 10^{-7}}{0.075^2}$

$E_2$ = $1.28 * 10^6 N/C$

The net electric field will be:

E = $E_1$ + $E_2$

E = $-1.28 * 10^6$ + $1.28 * 10^6$

E = 0 N/C

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3 years ago

An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid

Katena32 [7]

Answer:

666000000000 J

391764.71 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

Change in energy

$\Delta KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow \Delta KE=\frac{1}{2}3.7\times 10^4(4500^2-7500^2)\\\Rightarrow \Delta KE=-666000000000\ J$

Work done be the force is 666000000000 J

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{-666000000000}{1.7\times 10^6}\\\Rightarrow F=-391764.71\ N$

The magnitude of force is 391764.71 N

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Answer:

gravitional pull

Explanation:

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2 years ago