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1 year ago

New seafloor material is formed at places in Earth’s crust where –

Physics

2 answers:

mr_godi [17]1 year ago

4 0

Answer:

D, I think.

Explanation:

I had a quiz in Plate Tectonics and there was 2 questions that are related to this, but not the exact question.

Which material rises from cracks in oceanic crust

-molten rock

Which is the first step in the seafloor spreading process?

-a crack forms in oceanic crust.

those are all right btw, so you can decide if the answer I told you is right or not.

nlexa [21]1 year ago

3 0

Answer:

Explanation:

Give the other person brainliest he has a good explanation.

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A particular light source gives off light waves with a measured wavelength of

Umnica [9.8K]

The frequency of the light source is 1.5 x 10¹⁵ Hz.

<h3>Frequency of the light source</h3>

The frequency of the light source is determined using the following equations;

c = fλ

where;

c is speed of light

f is the frequency

λ is the wavelength

f = (3 x 10⁸) / (2 x 10⁻⁷)

f = 1.5 x 10¹⁵ Hz

Thus, the frequency of the light source is 1.5 x 10¹⁵ Hz.

Learn more about frequency of light here: brainly.com/question/10728818

3 0

2 years ago

In a thunderstorm, the thunder occurs at the same time as lightning. Explain how does a person hears the thunder 5s after he see

sergey [27]

Answer:

If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you'll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close. Keep in mind that you should be in a safe place while counting.

Explanation:

4 0

2 years ago

Name two objects that have a high density.

Sonbull [250]

Answer:

Iron and stone

Explanation:

4 0

3 years ago

The following is current scientific evidence supporting the nebular theory on the formation of the solar system. the composition

andreyandreev [35.5K]

All planets orbit the sun in a plane, all the planets orbit the sun in the same direction, most of the planets rotate in the same direction. I'm not sure when and answer ends or begins on your question so you can choose from some of the answers I gave you.

6 0

3 years ago

Read 2 more answers

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

2 years ago

Read 2 more answers