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3 years ago

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Select all that apply. which of the following astronomers supported the sun-centered system? tycho brahe johannes kepler coperni

cus ptolemy

2 answers:

ser-zykov [4K]3 years ago

8 0

Answer:Ptolemy

Explanation; He supported heloiocentrism.

Which say the sun is in the middle

Alex73 [517]3 years ago

3 0

<h3 />

-Tycho Brahe and

-Ptolemy

<h3 />

The sun is a hot ball of glowing gases which is a star whose gravity holds the solar system together and also keeping all the planet and smallest particles of debris in its orbit.

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M+2-4=3 solve and get 50 pts

Katen [24]

Hey!

First, let's write the problem.
$M+2-4=3$
Subtract the numbers, we would do the following operation, $2-4=-2$
$M-2=3$
Add 2 to both sides.
$M-2+2=3+2$

This tells us that our final answer would be,
$M=5$

Thanks!
-TetraFish

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3 years ago

Read 2 more answers

At what time of year is the intensity of solar radiation striking each of earth's hemispheres weakest?

bagirrra123 [75]

<span>During winter for a given hemisphere, solar radiation reaches the lowest period of its annual cycle due to the tilt of the earth on its axis. As the earth rotates around the sun, this tilt occludes a portion of the energy released by the sun as it diffuses in the atmosphere.</span>

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3 years ago

A car traveled 112.0m in 24.500s, what was a car’s average speed?

kodGreya [7K]

Average speed of the car is 4.57 m/s

Explanation:

Speed is calculated by the rate of change of displacement.

It is given by the formula, Speed = Distance/Time

Here, distance = 112 m and time = 24.5 s

Speed of the car = 112/24.5 = 4.57 m/s

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3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

Why is the law of gravity a scientific law

Yuri [45]

An object in motion will stay in motion unless acted upon another force.

Newton used this to prove that gravity existed. Without an unseen force, we could throw a ball and it would go on forever correct? Unless there was something to pull it down, in this case, gravity.

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3 years ago