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makvit [3.9K]
3 years ago
8

Gli iceberg nel Nord Atlantico rappresentano un rischio per la navigazione, costringendo ad allungare le rotte di navigazione di

circa il 30% durante la stagione della loro presenza. I tentativi per distruggere gli iceberg comprendono l’inserimento di esplosivi, bombardamenti, siluramenti, speronamenti, rivestimenti con fuliggine nera. Supponete che si sia provato a fondere direttamente l’iceberg, ponendo delle sorgenti calde nel ghiaccio.
Quanto calore è necessario per fondere il 10% di un iceberg da 200 000 tonnellate?
Scrivere il calcolo per esteso, non solo risultato numerico.
Physics
1 answer:
nikklg [1K]3 years ago
6 0

I DONT UNDERSTND IS ANOTHER LANGUAGR

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Bohr’s atomic model differed from Rutherford's because it explained that electrons exist in specified energy levels surrounding
insens350 [35]

Answer:

electrons exist in specified energy levels

Explanation:

In its gold-foil scattering with alpha particles, Rutherford proved that the plum-pudding model of the atom theorised by Thomson was wrong.

From his experiment, Rutherford inferred that the atom actually consists of a very small nucleus, where all the positive charge is concentrated, and the rest of the atom is basically empty, with the electrons (negatively charged) orbiting around the nucleus at very large distance.

However, Rutherford did not specify anything about the orbits of the electrons. Later, Bohr predicted that the electrons actually orbit the nucleus in specific orbits, each orbit corresponding to a specific energy level. Bohr's model found confirmation in the observation of the emission spectrum lines: when an electron in one of the higher energy level jumps down into an orbit with lower energy, the atom emits a photon which has an energy exactly equal to the difference in energy between the two orbits (and this energy of the photon corresponds to a precise wavelength).

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3 years ago
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What is 1113.28 in a sig fig?
Afina-wow [57]

Answer:

6 significant figure

Explanation:

The digits 111328 all are 6 figures with no figure being zero, neither zero after the other digits. In this case, all the numbers are significant and since they are only six numbers, then this is a six significant figure. In case we add another zero after digit 8, the zero is not significant but if added either infront of 8 or 2, the zero becomes significant.

4 0
3 years ago
How many earths can fit on the moon
liberstina [14]
Hello My Dear Friend!

Do you mean Sun or actually Moon?

If Moon is what you're looking for...then that is absolutely IMPOSSIBLE, due to that the Earth is WAY bigger than the Moon, therefore, not even HALF of the Earth would fit into the Moon since it's too small XD.

But if it's Sun...well then, t<span>he Answer is that it would take </span>1.3 million Earths<span> to fill up the Sun. That's a lot of Earths LOL XD

</span>I Hope my answer has come to your Help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead! :)

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8 0
3 years ago
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
3 years ago
Write the first equation of motion. Under what condition(s) is this equation valid?​
Zepler [3.9K]

Explanation:

The first equation of motion in kinematics is given by :

v=u+at .....(1)

u is initial speed

a is acceleration

v is final speed

t is time

Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.

3 0
3 years ago
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