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3 years ago

What did ancient astronomers think areas of the moon called mares might be?

Physics

1 answer:

Bumek [7]3 years ago

4 0

The ancient astronomers think areas of the moon called mares might be Seas.

Option D

Explanation:

The surface area of Earth's moon is dark, large, and is basaltic plains which are formed by ancient volcanic eruptions. They were dubbed as Maria, "ancient astronomers" who misunderstood them as actual seas. They are less reflective than highlands. Due to their iron-rich composition, they tend to appear dark from the naked eye. The Maria cover approximately about 16% of surface mostly on side that is visible from Earth. The few Maria on side that is too far are much smaller and residing mostly in very large craters. The ancient astronomers mistook the surface area as look like actual seas.

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Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

Serggg [28]

Answer:

(a) The force between them quadruples

Explanation:

According to coulomb's law, initial force between the two charged objects is given as;

$F_1=\frac{Kq_1q_2}{r^2}$

where;

k is coulomb's constant

q₁ is the charge on the first object

q₂ is the charge on the second object

r is the distance between the two objects

When the charges on both objects are doubled, then;

q₁ = 2q₁

q₂ = 2q₂

Force between the two charged objects will become

$F_2 = \frac{K2q_12q_2}{r^2} = \frac{4Kq_1q_2}{r^2} = 4(\frac{Kq_1q_2}{r^2}) = 4F_1$

Therefore, the force between them quadruples

4 0

3 years ago

11) In which state of matter has the LEAST kinetic energy?

klio [65]

Solid particles hope this helped!

5 0

2 years ago

Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is

Llana [10]

Answer:

$K_e_q=22.75878093\frac{N}{m}$

$f=1.363684118Hz$

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

$\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}$

Replacing the data provided:

$\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}$

$K_e_q=22.75878093\frac{N}{m}$

Finally, to calculate the frequency of oscillation we use this:

$f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }$

Replacing m and k:

$f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz$

4 0

3 years ago

Protons in an atomic nucleus are typically 10−15 m apart. what is the electric force (in n) of repulsion between nuclear protons

dybincka [34]

The electric force is given by:
F = [ k*(q1)*(q2) ] / d^2
F = Electric force
k = Coulomb's constant
q1 = Charge of one proton
q2 = Charge of second proton
d = Distance between centers of mass
Values:
F = unknown
k = 8.98E 9 N-m^2/C^2
q1 = 1.6E-19
q2 = 1.6E-19
d = 1.0E-15 m
Insert values into F = [ k*(q1)*(q2) ] / d^2
F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2
F = 229.888 N
answer
the electric force of repulsion between nuclear protons is 229.888 N

3 0

3 years ago

Read 2 more answers

Would a ship in Lake Ontario (fresh water) float higher or lower in the water than in the Atlantic Ocean (salt water)?

ohaa [14]

Atlantic Ocean (salt water)

8 0

2 years ago