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scoundrel [369]
2 years ago
14

The comet-hunting astronomer who made a list of over 100 nebulae and galaxies that could be mistaken for comets was Group of ans

wer choices
Physics
1 answer:
Keith_Richards [23]2 years ago
5 0

An Astronomer who made a list of 100 nebulae and galaxies that could

be mistaken for comets was Charles Messer's.

Charles Messier's was a very famous scientist in the field of Astronomy.

His research in appearance of Great comet filled spark to his passion

 of Astronomy. On January 21, 1759 he put his thought of wrong

 calculation of Delisle's and he described a faint glow resembling comet

 had been seen earlier. Then after, he discovered great comet near

 sword of Orion.

  There are various tales of Charles Messiers which put him forward

  towards his love of Astronomy.

        Learn more about comet and Nebulae here:

            brainly.com/question/21420531

                #SPJ4

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Describe how the coriolis effect acts differently based on location on earth
sammy [17]

Explanation:

The Coriolis effect happens when an entity is perceived from a moving reference frame going in a straight path. The changing reference frame makes the object appear as if it were moving along a curved road.

Circulation is counter-clockwise in the northern hemisphere. Circulation is clockwise in the southern hemisphere, and it is the equator, it is straight down without circulation.

6 0
3 years ago
On a summer afternoon, the sand on the beach can get very hot. When you step on the sand in bare feet, you can burn yourself.
8_murik_8 [283]
<span>Heat from the Sun is transferred to the sand without direct contact. This heat is then transferred to your feet by direct contact.</span>
8 0
3 years ago
Read 2 more answers
The Burj Khalifa in Dubai is the world's tallest building. The structure is 828 m (2716.5 ft) and has more than 160 stories. A t
lakkis [162]

Explanation:

Below is an attachment containing the solution.

5 0
3 years ago
Problem A spring was at its resting position where it is attached to a wall at its left side and a block at its right side as sh
puteri [66]

Answer:

F = 19.1 N

Explanation:

To find the force exerted by the string on the block you use the following formula:

F=kx  (1)  

k: spring constant = 95.5 N/m

x: displacement of the block from its equilibrium position = 0.200 m

you replace the values of k and x in the equation (1):

F=(95.5N/m)(0.200m)=19.1N

Hence, the force exterted on the block is 19.1 N

 

8 0
3 years ago
What fraction of 5 MeV α particles will be scattered through angles greater than 8.5° from a gold foil (Z = 79, density = 19.3 g
aalyn [17]

Answer:

f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)

And after do all the operations we got:

f(8) = 1.96 x10^{-4] atoms/m^2

And that would be the final answer for this case.

Explanation:

For this case we can use the fomrula for the fraction of incident particles scattered by an angle \theta, given by:

f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)

Where:

Z_1 e represent the charge of the projectile (Z1=2)

Z_2 e is the target charge (z2=79)

K= 5x10^6 eV represent the kinetic energy of incident particle

n represent the density of target particles (we need to find it first)

t= 10^{-8] m represent the thicknss of the foil

The first step would be calculate the density of target particles with the following formula:

n =\frac{\rho N_A N_M}{M_g}

Where:

\rho = 19.3 g/m^3= 19300 Kg/m^3

N_A = 6.022 x10^{23} molecules/mol the Avogadro's number

N_M = 1 represent the atoms per molecule

M_g = 197 g/mol = 0.197 Kg/mol represent the molecular weigth

If we replace we got:

n = \frac{19300 kg/m^3 *6.022x10^{23} molecu/mol * 1 atom/mole}{0.197 Kg/mol}= 5.90 x106{28] atoms/m^3

Now we can calculate the fraction of 5 MeV alpha particles that would be scatteres with angle higher than 8 degrees in a piece of thickness t=10^{-8}m

And using the first formula we got:

f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)

And after do all the operations we got:

f(8) = 1.96 x10^{-4] atoms/m^2

And that would be the final answer for this case.

4 0
3 years ago
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