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3 years ago

What exactly does it mean to say an object is polarized

1 answer:

5 0

It means that it has a force to make something go in one directions only
if a metal is polarized it means it has a pith pole and a north pole like a magnet

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Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude

natka813 [3]

Answer:

For vector u, x component = 10.558 and y component =12.808

unit vector = 0.636 i+ 0.7716 j

For vector v, x component = 23.6316 and y component = -6.464

unit vector = 0.9645 i-0.2638 j

Explanation:

Let the vector u has magnitude 16.6

u makes an angle of 50.5° from x axis

So $u_x=ucos\Theta =16.6\times cos50.5=10.558$

Vertical component $u_y=usin\Theta =16.6\times sin50.5=12.808$

So vector u will be u = 10.558 i+12.808 j

Unit vector $u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j$

Now in second case let vector v has a magnitude of 24.5

Making an angle with -15.3° from x axis

So horizontal component $v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316$

Vertical component $v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464$

So vector v will be 23.6316 i - 6.464 j

Unit vector of v $=\frac{23.6316i-6.464}{\sqrt{23.6316^2+6.464^2}}=0.9645i-0.2638j$

8 0

2 years ago

The image shows street lights powered by solar panels. Which sequence shows the energy transformations taking place in these lig

yawa3891 [41]

the answer is C. for plato

8 0

2 years ago

Read 2 more answers

A constant torque is applied to a rigid wheel whose moment of inertia is 2.0 kg · m2 around the axis of rotation. If the wheel s

Jlenok [28]

Answer:

The applied torque is 3.84 N-m.

Explanation:

Given that,

Moment of inertia of the wheel is $2\ kg-m^2$

Initial speed of the wheel is 0 (at rest)

Final angular speed is 25 rad/s

Time, t = 13 s

The relation between moment of inertia and torque is given by :

$\tau=I\alpha \\\\\tau=I\times \dfrac{\omega_f}{t}\\\\\tau=2\times \dfrac{25}{13}\\\\\tau=+3.84\ N-m$

So, the applied torque is 3.84 N-m.

4 0

2 years ago

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

4 0

2 years ago

If 2N force is applied on 2 kg mass due east and same magnitude of force due west, thechange in velocity of the body in 2 sec is

klio [65]

Explanation:

F=m(v-u)/t

F=2N

m=2kg

t=2s

2=2(v-u)/2

cross multiply

2*2=2(v-u)

4=2(v-u)

4/2=v-u

v-u=2m/s

v-u is the change is velocity.

3 0

2 years ago