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3 years ago

Which describes the motion of the media for a surface wave?

Physics

1 answer:

Sidana [21]3 years ago

4 0

Answer:

a. the motion is both parallel and perpendicular

Explanation:

I did grad point :)

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An electron initially 3.00 m from a nonconducting infinite sheet of uniformly distributed charge is fired toward the sheet. The

Cloud [144]

Answer:

$4.4443704375\times 10^{-18}\ C/m^2$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

$\Delta l$ = Distance charge traveled = 2 m

v = Velocity of electron = 420 m/s

E = Electric field

$m_e$ = Mass of electron = $9.11\times 10^{-31}\ kg$

$q_e$ = Charge of electron = $1.6\times 10^{-19}\ C$

As the energy of the system is conserved we have

$q_eE\Delta l=\dfrac{1}{2}m_ev^2\\\Rightarrow E=\dfrac{1}{2}\dfrac{m_e}{q_e}\times \dfrac{v^2}{\Delta l}\\\Rightarrow E=\dfrac{1}{2}\dfrac{9.11\times 10^{-31}}{1.6\times 10^{-19}}\times \dfrac{420^2}{2}\\\Rightarrow E=2.51094375\times 10^{-7}\ N/C$

For an infinite non conducting sheet electric field is given by

$E=\dfrac{\sigma}{2\epsilon}\\\Rightarrow \sigma=2E\epsilon\\\Rightarrow \sigma=2\times 2.51094375\times 10^{-7}\times 8.85\times 10^{-12}\\\Rightarrow \sigma=4.4443704375\times 10^{-18}\ C/m^2$

The surface charge density is $4.4443704375\times 10^{-18}\ C/m^2$

5 0

3 years ago

You buy a 75 watt light bulb. the label means that

alukav5142 [94]

The label means that the bulb will consume 75 joules of electrical energy in 1 sec

4 0

4 years ago

An object is launched from the top of a building which is 100 m tall (relative to the ground) at a speed of 22 m/s at an angle o

klio [65]

Answer:

$3.58\:\mathrm{s}$

Explanation:

We can use the kinematics equation $\Delta y=v_it+\frac{1}{2}at^2$ to solve this problem. To find the initial vertical velocity, find the vertical component of the object's initial velocity using basic trigonometry for right triangles:

$\sin28^{\circ}=\frac{y}{22},\\y=22\sin28^{\circ}=10.3283743813\:\mathrm{m/s}$

Now we can substitute values in our kinematics equation:

$\Delta y=-100$
$a=-9.8\:\mathrm{m/s^2}$ (acceleration due to gravity)
$v_i=-10.3283743813\:\mathrm{m/s}$
Solving for $t$

$-100=-10.3283743813t+\frac{1}{2}\cdot -9.8\cdot t^2,\\\\-4.9t^2-10.3283743813t+100=0,\\\\\boxed{t=3.5849312673637455}, t=-5.692762773751501\:\text{(Extraneous)}$