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Alexxx [7]
3 years ago
7

Which shows the weight of an atom?

Physics
2 answers:
harina [27]3 years ago
8 0

Answer:

Atomic mass

Explanation:

Dmitry_Shevchenko [17]3 years ago
7 0

Answer:

atomic mass

Click thanks if thankc if this helped

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What evidence can you cite to support the claim that the frequency of light does not change upon reflection?
soldier1979 [14.2K]

Answer: The color of an image is equal to the color of the item forming the picture. When you have a take a observe your self in a mirror, the color of your eyes would not alternate. The reality that the color is equal is proof that the frequency of light would not alternate upon reflection.

8 0
4 years ago
The Escape speed at the surface of a certain planet is twice that of the earth. what is its mass in unit of earth's mass?
VLD [36.1K]

Answer:

22Km/sec

Explanation:

6 0
3 years ago
كرة كتلتها و 200 مربوطة بنهاية خيط طوله m 1 تتحرك حركة دائرية منتظمة في مسار دائري
Vesna [10]
Asalam a lekum my brother
3 0
3 years ago
When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to
Serjik [45]

Answer:

E) d/sqrt2

Explanation:

The initial electric force between the two charge is given by:

F=k\frac{q_1 q_2}{d^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

d is the separation between the two charges

We can also rewrite it as

d=\sqrt{k\frac{q_1 q_2}{F}}

So if we want to make the force F twice as strong,

F' = 2F

the new distance between the charges would be

d'=\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{1}{\sqrt{2}}\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{d}{\sqrt{2}}

so the correct option is E.

8 0
3 years ago
A 23 g bullet is accelerated in a rifle barrel 62 cm long to a speed of 593 m/s. Use the work-energy theorem to find the average
Anuta_ua [19.1K]

Answer:

The average force exerted on the bullet while it is being accelerated is 6,522.52 N.

Explanation:

Given;

mass of the bullet, m = 23 g = 0.023 kg

length of the barrel, L = 62 cm = 0.62 m

speed of the bullet, v = 593 m/s

Applying work-energy theorem;

the work done in accelerating the bullet in the riffle = kinetic energy acquired by the bullet.

W = K.E

F x d = ¹/₂mv²

where;

d is the is the distance traveled by the bullet in the riffle = L

F(0.62) = ¹/₂ x 0.023 x (593²)

F(0.62) = 4043.964

F = (4043.964) / (0.62)

F = 6,522.52 N

Therefore, the average force exerted on the bullet while it is being accelerated is 6,522.52 N.

6 0
3 years ago
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