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3 years ago

What are somethings about the Heliocentric model?

Physics

1 answer:

Rudik [331]3 years ago

6 0

Answer:

Heliocentrism, a cosmological model in which the Sun is assumed to lie at or near a central point (e.g., of the solar system or of the universe) while the Earth and other bodies revolve around it.

Explanation:

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A 5.6 g marble is fired vertically upward using a spring gun. The spring must be compressed 6.4 cm if the marble is to just reac

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Answer: im not gonna give i to you just do 15+15=_+ 5.6+6.4 easy

Explanation: i took the test and got a 100%

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2 years ago

If a one-pound weight were one foot from the pivot or shaft of a lever, how many pound-feet of force would result? A. 4 B. 1 C.

mixas84 [53]

The force is still 1 pound.

The TORQUE around the pivot is (force) x (distance from the pivot) = 1 foot-pound. (B)

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3 years ago

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A 193nm-wavelength UV laser for eye surgery emits a 0.500mJ pulse. (a) How many photons does the light pulse contain?

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Using the equation E = hc/λ we can find out how much energy a single photon of wavelength 193 nm has. E = Planck Constant * Speed of Light/193 nm

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3 years ago

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Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

$F = G \frac{m_1m_2}{r^2}$

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

$F = m_2a$

Therefore, we can see that

$a(r) = G \frac{m_1}{r^2}$

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}$

a(R) is actually the constant acceleration at sea level

and

$a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}$

Therefore:

$a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})$

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}$

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3 years ago

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The Periodic Table of Elements is organized by the number of:

QveST [7]

The answer to your question is Protons!

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