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mixer [17]
2 years ago
13

As crust forms at the mid-ocean ridge, it gets pushed away from the ridge as newer crust forms. What is true about crust that fo

rms at the mid-ocean ridge?
A: crust further away from the mid-ocean ridge is younger and crust closer is older


B: crust further away from the mid-ocean ridge is older and crust closer is newer


C: crust is not created at the mid-ocean ridge


D: crust further away from the mid-ocean ridge and crust closer are the same age
Physics
2 answers:
MrRa [10]2 years ago
4 0
The to the question is D
Nuetrik [128]2 years ago
3 0

Answer:

Answer is A

Explanation:

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What will a spring scale read for the weight of a 57.0-kg woman in an elevator that moves upward with constant speed of 5.0 m/s
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The spring scale will read 559 Newton's or 125.7 pounds.
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3 years ago
A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is
puteri [66]

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}

\dfrac{1}{u}=\dfrac{103}{4080}

u =\dfrac{4080}{103}

The magnification is

m = \dfrac{-v}{u}

m=\dfrac{-240\times103}{4080}

m = -6.05

Hence, The magnification is -6.05.

6 0
2 years ago
Choose all that apply. Solids, liquids, and gases can be distinguished by their:
vodomira [7]
I believe its by there shape
8 0
3 years ago
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Which phase of the Moon occurs when Earth, the Moon, and the Sun lie on the same line?
ss7ja [257]
The answer is B I think.
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3 years ago
The Franck-Hertz experiment involved shooting electrons into a low-density gas of mercury atoms and observing discrete amounts o
Anuta_ua [19.1K]

Answer:

the final kinetic energy is 0.9eV

Explanation:

To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:

E_n=\frac{-13.6eV}{n^2}

you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is

E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV

-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:

E_{k}=11.1eV-10.2eV=0.9eV

6 0
3 years ago
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