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Mariulka [41]
3 years ago
8

When rocks are analyzed over an increasingly greater distance from a mid-ocean ridge, what can be said about the polarities of t

he rocks?
Physics
1 answer:
zubka84 [21]3 years ago
5 0
They are of more, or opposite polaritys
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98 POINTS FOR ANYONE THAT DOES THIS NOW!!
Varvara68 [4.7K]

In simpler terms, a proton and neutron weigh 1 amu (atomic mass unit) each.

The nucleus has 15 protons and 18 neutrons. Since a proton's and a neutron's weight is only 1 amu, we can simply add the number of protons and neutrons to find the total mass of the nucleus:

15 + 18 = \boxed{33}

The nucleus' mass is 33 amu.

8 0
3 years ago
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Most weather occurs in which atmospheric layers?
Slav-nsk [51]

Answer:

Troposphere

Explanation:

The troposphere is the lowest layer of the Earth's atmosphere. The earth's atmosphere has the following layers:

- Troposphere - Stratosphere - Mesosphere -Thermosphere - Exosphere

The troposphere lies 10 km above the land surface and most weather phenomena and conditions (e. g. clouds, turbulence) occur there.

Most of the water vapor in the atmosphere is in the troposphere (about 99%) and most of the atmosphere's mass is contained in the troposphere (75%)

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4 years ago
A solid sphere of radius R carries a fixed, uniformly distributed charge q. Obtain an expression for the magnitude of the electr
NISA [10]

Answer:

The electric field outside the sphere will be \dfrac{qr}{4\pi\epsilon_{0}R^3}.

Explanation:

Given that,

Radius of solid sphere = R

Charge = q

According to figure,

Suppose r is the distance between the point P and center of sphere.

If \rho be the volume charge density,

Then, the charge will be,

q=\rho\times\dfrac{4}{3}\pi R^3.....(I)

Consider a Gaussian surface of radius r.

We need to calculate the electric field outside the sphere

Using formula of electric field

\oint{\vec{E}\cdot \vec{dA}}=\dfrac{Q}{\epsilon_{0}}

E\times4\pi r^2=\dfrac{\rho\dotc \dfrac{4}{3}\pi r^3}{\epsilon_{0}}

Put the value from equation (I)

E\times4\pi r^2=\dfrac{qr^3}{\epsilon_{0}R^3}

E=\dfrac{qr}{4\pi\epsilon_{0}R^3}

Hence, The electric field outside the sphere will be \dfrac{qr}{4\pi\epsilon_{0}R^3}.

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3 years ago
I feel dvmb for asking this question but... help? Please?
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<u>Force</u>, it can speed up or slow down an object which can change the direction in which an object is moving.

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HELP PLEASE!!! I'M REALLY BAD AT MECHANICAL ENERGY!!!
deff fn [24]
Kinetic is something moving and potential is being still

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