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3 years ago

Find an expression for the square of the orbital period.

Physics

1 answer:

jenyasd209 [6]3 years ago

8 0

Answer:

T²= 4π²R³/GM

Explanation:

First we know that

Fg= Fc

Because centripetal force must equal gravitational force

GMm/R² = Mv²/R

But velocity is 2πR/T

So by substitution we have

GMm/R²= M (2πR/T)/T

We have

T²= 4π²R³/GM as period

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On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophou

Vilka [71]

The period of the pendulum is given by the following equation

T = 2π * sqrt (L/g)

Where g is the gravity (free fall acceleration)

L is the longitude of the pendulum

T is the period.

We find g.............> (T /2π)^2 = L/g

g = L/(T /2π)^2...........> g = 22.657 m/s^2

3 0

3 years ago

On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of

Advocard [28]

Answer:

On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of electrons? (the mass will decrease by a very small factor)

(b) What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons? (The mass will increase by a very small factor)

Explanation:

(a) On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of electrons? (the mass will decrease by a very small factor)

The mass of an atom is given by the sum of the masses of the protons, neutrons and electrons. Electrons has lower mass than protons and neutrons, so they have a minor contribution to the total mass of the atom.

When an object is electrically neutral it means that it has the same number of protons and electrons. For the case of an object positively charged, the rate of protons is greater than the number of electrons. That means that atom lose electrons so the mass will decrease in a very small factor.

(b) What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons? (The mass will increase by a very small factor)

For the case when the object is negatively charged, it means that the atom gains electrons from another object, leading to the conclusion that the mass of the atom will increase in a very small factor.

Key values:

Electron mass: 9.1095×10⁻³¹ Kg

Proton mass: 1.67261×10⁻²⁷ Kg

Neutron mass: 1.67492×10⁻²⁷ Kg

5 0

3 years ago

Water near the poles would most likely be stored as

solniwko [45]

Hey there!

Answer: Glaciers

Water near the poles would most likely be stored as glaciers. Glaciers are slow moving rivers that are a buildup of ice and snow.

Thank you!

5 0

3 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

3 years ago

What is the half-life of a radioisotope

IceJOKER [234]

The rate at which a radioactive isotope decays is measured in half-life. The termhalf-life is defined as the time it takes for one-half of the atoms of a radioactive material to disintegrate. Half-lives for various radioisotopes can range from a few microseconds to billions of years.

4 0

3 years ago

Read 2 more answers