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3 years ago

Convection currents in this layer drive the tectonic plates above them.

Physics

1 answer:

Sphinxa [80]3 years ago

7 0

Hello, hello? Hey! Hey, wow, day 4. I knew you could do it.

Uh, hey, listen, I may not be around to send you a message tomorrow. *banging* It's-It's been a bad night here for me. Um, I-I'm kinda glad that I recorded my messages for you *clears throat* uh, when I did.

Uh, hey, do me a favor. *banging* Maybe sometime, uh, you could check inside those suits in the back room? *banging* I'm gonna to try to hold out until someone checks. Maybe it won’t be so bad. *bang bang* Uh, I-I-I-I always wondered what was in all those empty heads back there. *chime plays*.

You know...*deep moan* oh, no - *noises followed by a loud screech and static*

Explanation:

You might be interested in

The courtroom role that possesses the most discretion is?

KonstantinChe [14]

The role of the Prosecutor's

5 0

4 years ago

Svetlanka [38]

Answer:

North and east are at 90 degree, equation becomes

Explanation:

(2x 2)+[2(x+5)] 2=50 2

on solving, we get

x=12.366km/h

4 0

3 years ago

What is the wavelength of a wave with a frequency of 262 hz and a speed of 343 m/s

Tamiku [17]

Heya!!

For calculate wavelength, lets applicate formula:

$\boxed{\lambda = V/f}$

<u>Δ Being Δ</u>

f = Frequency = 262 Hz

v = Velocity = 343 m/s

$\lambda$ = Wavelenght = ?

⇒ Let's replace according the formula:

$\boxed{\lambda = 343\ m/s / 262\ Hz }$

⇒ Resolving

$\boxed{\lambda = 1,3 \ m}$

Result:

The wavelength is <u>1,3 meters.</u>

Good Luck!!

8 0

3 years ago

Read 2 more answers

(1) Develop an equation that relates the rms voltage of a sine wave to its peak-to-peak voltage. a. If a sine wave has a peak-to

Aleks04 [339]

Answer:

(A) Equation will be $v=v_msin\omega t=0.75sin(18840t)$

(B) RMS value of voltage will be 0.530 volt

Explanation:

We have given peak to peak voltage of ac wave = 1.5 volt

Peak to peak voltage of ac wave is equal to 2 times of peak voltage

So $2v_{peak}=1.5volt$

$v_{peak}=\frac{1.5}{2}=0.75volt$

Frequency of ac wave is given f = 3 kHz

So angular frequency $\omega =2\pi f$ = 2×3.14×3000 = 18840 rad/sec

So expression of equation will be $v=v_msin\omega t=0.75sin(18840t)$ ( As phase difference is 0 )

Now we have to find the rms value of voltage

So rms voltage will be equal to $v_{rms}=\frac{v_{peak}}{\sqrt{2}}=\frac{0.75}{1.414}=0.530volt$

4 0

3 years ago

In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and part

lesya [120]

Answer:

Charge of particle 2, $q_2=-7.13\ \mu C$

Explanation:

Given that,

Charge 1, $q_1=3.11\ \mu C=3.11\times 10^{-6}\ C$

The distance between charges, r = 0.241 m

Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$q_2=\dfrac{Fr^2}{kq_1}$

$q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}$

$q_2=7.13\times 10^{-6}\ C$

$q_2=7.13\ \mu C$

So, the magnitude of electric charge 2 is $q_2=7.13\ \mu C$ . Since, the force is attractive then the magnitude of charge 2 must be negative.

5 0

3 years ago