what moon phase will occur about 14 days after a first quarter?

In which situation are waves transmitted?

marin [14]

Answer:

In electromagnetic waves.

Explanation:

Energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.

3 0

3 years ago

Read 2 more answers

as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p

miv72 [106K]

Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

$Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta$

where,

E = Magnitude of Electric Field = ?

v = speed of charge = 230 m/s

B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

Therefore,

$E = (3)(230\ m/s)(0.61\ T)Sin90^o$

E = 420.9 N/C

3 0

3 years ago

Constants Modern wind turbines are larger than they appear, and despite their apparently lazy motion, the speed of the blades ti

nadezda [96]

It is given that the length of blade of the turbine is 58 m.

During the motion, the turbine will undergo rotational motion. Hence the radius of the circle traced by the turbine is equal to the length of the blade.

Hence radius r = 58 m.

The frequency of the turbine [f] =14 rpm.

Here rpm stands for rotation per minute.

Hence the frequency of the turbine in one second-

$f=\frac{14}{60}\ s^-1$

$f=0.23333 Hz$

Here Hz[ hertz] is the unit of frequency.

The angular velocity of the turbine $\omega =2\pi f$

$\omega=2*3.14*0.2333$

$\omega=1.465124$ radian/second

Now we have to calculate the centripetal acceleration of the blade.

Let the linear velocity of the blade is v.

we know that linear velocity v=ωr

The centripetal acceleration is calculated as-

$a_{c} =\frac{v^2}{r}$

$=\frac{[\omega r]^2}{r}$

$=\omega^2r$

$=[1.465124]^2 *58$

$=124.5021234 m/s^2$ [ans]

8 0

3 years ago

The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequen

Eddi Din [679]

Answer:

$\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }$

Explanation:

Additional information:

The ball has charge $-q_0$ , and the ring has positive charge $+Q$ distributed uniformly along its circumference.

The electric field at distance $z$ along the z-axis due to the charged ring is

$E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.$

Therefore, the force on the ball with charge $-q_0$ is

$F=-q_oE_z$

$F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}$

and according to Newton's second law

$F=ma=m\dfrac{d^2z}{dz^2}$

substituting $F$ we get:

$- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}$

rearranging we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0$

Now we use the approximation that

$z^2+a^2\approx a^2$ (we use this approximation instead of the original $d^2+a^2\approx a^2$ since $z$ , our assumption still holds )

and get

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0$

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0$

Now the last equation looks like a Simple Harmonic Equation

$m\dfrac{d^2z}{dz^2}+kz=0$

where

$\omega=\sqrt{ \dfrac{k}{m} }$

is the frequency of oscillation. Applying this to our equation we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}$

$\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}$

5 0

4 years ago

The pressure at the bottom of a jug filled with water does NOT depend on the

pickupchik [31]

Answer:

D) surface area of the water

Explanation:

The pressure at the bottom of a column of fluid is given by Stevino's law:

$p=\rho g h$

where

p is the pressure at the bottom of the column

$\rho$ is the density

g is the acceleration due to gravity

h is the depth of the liquid

So, we see that the pressure at the bottom of a jug filled with water depends on all these quantities:

A) depth of the liquid

B) acceleration due to gravity

C) density of water

While it does not depend on:

D) surface area of the water

5 0

3 years ago

￼￼what moon phase will occur about 14 days after a first quarter?

what moon phase will occur about 14 days after a first quarter?