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denpristay [2]
4 years ago
11

Which layer(s) of Earth participate(s) in plate tectonics? Select all that apply.

Physics
2 answers:
mafiozo [28]4 years ago
8 0
I am pretty sure it is A and D
nadezda [96]4 years ago
5 0

A and D are the correct answers.

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Describe the relationship between the length and period of a pendulum in the language of direct proportions
Wittaler [7]

The period of the pendulum is directly proportional to the square root of the length of the pendulum

Explanation:

The period of a simple pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that when the length of the pendulum increases, the period of the pendulum increases as the square root of L, T\propto \sqrt{L}. This means that

The period of the pendulum is directly proportional to the square root of the length of the pendulum

From the equation, we also notice that the period of a pendulum does not depend on its mass.

#LearnwithBrainly

3 0
3 years ago
Another name for a period is a family?
Zepler [3.9K]
If you are referring to science, then you are correct. A period is also known as a family.
8 0
3 years ago
Photons with the highest energy have the ____
Setler79 [48]

Answer:

D. shortest wavelength

Explanation:

Photons with the highest energy have the shortest wavelength. The shorter the wavelength, the higher the energy of a photon.

A photon is a quantity that transmits electromagnetic energy from one place to the other.

  • Gamma rays have photons that transmits the highest amount of energy.
  • The rays have the shortest wavelength and highest frequency of all electromagnetic radiations.

Energy, wavelength and frequency of a photon are connected using the expression:

    E =  h f  = \frac{hc}{wavelength}  

E is the energy

h is the Planck's constant

 f is the frequency.

3 0
3 years ago
Alguien me ayuda con calculos estequiometricos?
ruslelena [56]

Answer:

SURE!!!...

But what to calculate!!!....

5 0
3 years ago
An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane
lubasha [3.4K]

Answer:

\rho = 12580.7 kg/m^3

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

F = \frac{GMm}{(r + h)^2}

here we have

F =\frac {mv^2}{(r+ h)}

\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}

here we have

v = \sqrt{\frac{GM}{(r + h)}}

now we can find time period as

T = \frac{2\pi (r + h)}{v}

T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}

1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}

M = 4.54 \times 10^{23} kg

Now the density is given as

\rho = \frac{M}{\frac{4}{3}\pi r^3}

\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}

\rho = 12580.7 kg/m^3

8 0
3 years ago
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