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3 years ago

11

A circular flat coil that has N turns, encloses an area A, and carries a current i, has its central axis parallel to a uniform m

agnetic field B in which it is immersed. The net force on the coil is
-Zero

-NiAB

-NiB

-NiA

-iBA

Is the answer zero because the coil is placed parallel to the magnetic field?

1 answer:

Aneli [31]3 years ago

4 0

Answer:

A. Zero

Explanation:

The force on a coil of N turns, enclosing an area, A and carrying a current I in the presence of a magnetic field B, is :

F = N * I * A * B * sinθ

Where θ is the angle between the normal of the enclosed area and the magnetic field.

Since the normal of the area is parallel to the magnetic field, θ = 0

Hence:

F = NIABsin0

F = 0 or Zero

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Explanation:

Given

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$\Delta E_2=61+W_2$

$\Delta E_1+\Delta E_2=0$

as the process is cyclic

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3 years ago

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli

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Answer:

Not possible

Explanation:

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$V_{\rho l}$ = Volume fraction of fibre (longitudinal)

$V_{\rho t}$ = Volume fraction of fibre (transvers)

$E_f$ = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

$E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764$

Transverse modulus of elasticity is given by

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3 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

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Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

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prohojiy [21]

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3 years ago

Read 2 more answers