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2 years ago

What confirms that the process of seafloor spreading is occurring?

1 answer:

3 0

Mid ocean ridges
The mountain ranges in the middle of the oceans
Sea floor spreading
The process by which new, oceanic crust forms along a mid ocean ridge and older crust moves away from the ridge. :)

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. Now assuming Anna's far point was found to be 0.9 m (i.e., her eyes can't focus on any object more than 0.9 m away), what powe

Mama L [17]

Answer:1). Distance of far point x=0.9m

Therefore, since the image is virtual

-f=-x = -0.9m

Power of the concave lenses = 1/f = 1/-0.9

= -1.11D

2 ) near point is 21cm = 0.21m

Power = 4-1/near point

= 4/0.21

= 14.2D.

7 0

3 years ago

which of the following objects is considered negatively charged A - one with excess electrons on the surface B - one with excess

Ne4ueva [31]

An object that could be considered as negatively charged would be when it has an excess of an electron in its atom. However, when it loses an electron, it could go back to its stable state which is "uncharged" or when there is an excess proton, it could be a positively charged object.

8 0

3 years ago

Which of the following statements are true about gravity? Check all that

mamaluj [8]

Answer:

c d and e

Explanation:

4 0

2 years ago

A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a sta

creativ13 [48]

Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

$F_c=F_e=ma_c$ (1)

m: mass of the particle = 20g = 20*10-3 kg

ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

$F_e=k\frac{qq'}{r^2}$ (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

$a_c=\frac{v^2}{r}$ (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

$k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}$

Finally, you replace the values of all parameters in the previous expression:

$r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m$

The radius of the circular trajectory is 0.22m

5 0

3 years ago

What is the magnetic force on a proton that is moving at 4.5 107 m/s to the right through a magnetic field that is 1.6 T and poi

pickupchik [31]

The answer is letter C. 1.2 10^-11 N up
Solution:
F= Bqvsin(theta)
theta = sin 90 = 1
F= 1.4 T * 1.6x10^-19 * 5.2x10^7 ms^-1
F= 1.16 x 10^-11 N
Then the direction is upward.

6 0

3 years ago

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