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3 years ago

True or false? The plates keep moving and are still moving today.

Physics

1 answer:

wariber [46]3 years ago

6 0

Answer:

true

Explanation:

takes time for most plates to move but throughout time they keep moving just at a slow pace

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The diagram shows the electric field lines surrounding two positive point charges. If the charge on the right were replaced with

WINSTONCH [101]

Answer:

Explanation:

The wording on some of these choices is very strange; I'm not sure exactly what they are stating. First of all, A. is definitely a choice because if both the charges were opposite, they would be attracted to one another as opposed to be repelled away from one another, as they are when they are both positive. What happens is that the charges go OUT from the positive charge and INTO the negative; so as far as the field lines around both charges would change direction...no; only the direction of the field lines would change on the positive charge (which is the one on the left). In that space where D is filled in by the field lines going OUT of the positive charge and INTO the negative one, the lines there are naturally closer together, and that is the point where the charge is the greatest. So if that is what is meant by the field lines getting closer together, then yes, they do. As far as choice D. again the field lines on the negative charge don't change, only the ones on the positive charge change.

4 0

3 years ago

A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 8.00î m/s. Find (a

sergejj [24]

a. The particle has position vector

$\vec r(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)t\right)\,\vec\imath+\left(\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\right)\,\vec\jmath$

$\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

b. Its velocity vector is equal to the derivative of its position vector:

$\vec v(t)=\vec r'(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath$

c. At $t=7.00\,\mathrm s$ , the particle has position

$\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath$

$\vec r(7.00\,\mathrm s)=\left(56.0\,\vec\imath+49.0\,\vec\jmath\right)\,\mathrm m$

That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of $\|\vec r(7.00\,\mathrm s)\|=\sqrt{(56.0\,\mathrm m)^2+(49.0\,\mathrm m)^2}=74.4\,\mathrm m$ away from the origin in a direction of $\theta=\tan^{-1}\dfrac{49.0\,\mathrm m}{56.0\,\mathrm m}=41.2^\circ$ relative to the positive $x$ axis.

d. The speed of the particle at $t=7.00\,\mathrm s$ is the magnitude of the velocity at this time:

$\vec v(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)\,\vec\jmath$

$\vec v(7.00\,\mathrm s)=\left(8.00\,\vec\imath+14.0\,\vec\jmath\right)\dfrac{\rm m}{\rm s}$

Then its speed at this time is

$\|\vec v(7.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+(14.0\dfrac{\rm m}{\rm s}\right)^2}=16.1\dfrac{\rm m}{\rm s}$

4 0

3 years ago

The sound emitted by bats has a wavelength of 4.13 mm.

blagie [28]

Answer:

the frequency of a wave is equal to the linear speed divided the wavelength. so in equation form.

f = v / l

where f is the frequency

v iss the linear speed

l is the wavelength

f = ( 345 m/s ) / ( 4.38 mm ) ( 1 m / 1000 mm )

f = 78767 per second

Explanation:

hope it helps

8 0

4 years ago

What two types of scientific knowledge can be expressed as mathematical equations? (Please don't tell me the answer could you pl

gtnhenbr [62]

An example would be 2 types of motion. It could be rectilinear or projectile motion. There are various equations for each type. Since you don't want me to tell you the answer, I could just express it in words. Then, it will be up to you to translate into mathematical equations.

For rectilinear motion, the distance traveled is equal to the initial velocity times the time, plus one-half of the acceleration times the square of the time. For projectile motion, the maximum distance is equal to the square of the initial velocity multiplied with the square of the sine of the launch angle, all over twice the gravity.

5 0

4 years ago

Two identical positive charges exert a repulsive force of 6.4x10^-9 N when separated by a distance of 3.8x10^10 m. Calculate the

aleksley [76]

Answer:F = kq2/d2 ⇒

q = √(Fd2/k)

q = d √(F/k)

d = 3.8 x 10-10 m

F = 6.4 x 10-9 N

Look up k in your physics book in appropriate units, and plug in the numbers. You should get q in coulombs.

Explanation:

8 0

3 years ago