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gizmo_the_mogwai [7]

3 years ago

10

An electric field of intensity 3.7 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.3

50 m wide and 0.700 m long if the following conditions are true:
a. The plane is parallel to the yz-plane.
___________N·m2/C
b. The plane is parallel to the xy-plane.
___________N·m2/C
c. The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.
___________N·m2/C

2 answers:

suter [353]3 years ago

7 0

Answer:

a) $906.5\ N .m^2/C$

b) $0\ N . m^2/C$

c) $742.5\ N . m^2/C$

Explanation:

Area of the plane is,

a) The plane is parallel to the yz-plane.

$A = (0.350)(0.700)\\= 0.245\ m^2\\\phi_{E} = EA \cos \theta \\= (3.70 \times 10^3)(0.245) \cos0^{\circ}\\\phi_{E} = 906.5\ Nm^2/C\\$

b) The plane is parallel to the x-axis, the normal line of the area in at a right angle

$\theta = 90^{\circ}\\ \phi_{E} = EA \cos \theta\\ \phi_{E} = (3.70 \times 10^3)(0.245) \cos 90^{\circ}\\ \phi_{E} = 0\ N . m^2/C\\$

c) The plane contains the y-axis, and its normal makes an angle of $35.0^{\circ}$ with the x-axis $35.0^{\circ}$

$\phi_{E} = EA \cos \theta\\ \phi_{E} = (3.70 \times 10^3)(0.245) \cos 35^{\circ}\\ \phi_{E} = 742.5\ N . m^2/C$

jekas [21]3 years ago

4 0

Answer:

a)906.5 Nm^2/C

b) 0

c) 742.56132 N•m^2/C

Explanation:

a) The plane is parallel to the yz-plane.

We know that

flux ∅= EAcosθ

3.7×1000×0.350×0.700=906.5 N•m^2/C

(b) The plane is parallel to the xy-plane.

here theta = 90 degree

therefore,

0 N•m^2/C

(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.

therefore, applying the flux formula we get

3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C

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B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

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

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

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