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barxatty [35]
3 years ago
5

Cuántas veces es mayor la masa del protón que la del electrón?

Physics
1 answer:
viktelen [127]3 years ago
6 0

Answer:

Un protón es aproximadamente 1835 veces más masivo que un electrón. Si está preguntando acerca de sus dimensiones físicas, nadie lo sabe. Los científicos actualmente no saben qué tan pequeños son los electrones. ¡Son más pequeños de lo que podemos medir actualmente y pueden no tener un tamaño en absoluto!

Explanation:

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The block has maximum kinetic energy at the bottom of the curved incline. Since its radius is 3.0 m, this is also the block's starting height. Find the block's potential energy <em>PE</em> :

<em>PE</em> = <em>m g h</em>

<em>PE</em> = (2.0 kg) (9.8 m/s²) (3.0 m)

<em>PE</em> = 58.8 J

Energy is conserved throughout the block's descent, so that <em>PE</em> at the top of the curve is equal to kinetic energy <em>KE</em> at the bottom. Solve for the velocity <em>v</em> :

<em>PE</em> = <em>KE</em>

58.8 J = 1/2 <em>m v</em> ²

117.6 J = (2.0 kg) <em>v</em> ²

<em>v</em> = √((117.6 J) / (2.0 kg))

<em>v</em> ≈ 7.668 m/s ≈ 7.7 m/s

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3 years ago
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3 years ago
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3. What is the relationship among wave speed, wavelength, and frequency?
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Answer:

Wave speed is related to both wavelength and wave frequency. Wavelength is the distance between two corresponding points on adjacent waves. Wave frequency is the number of waves that pass a fixed point in a given amount of time. This equation shows how the three factors are related:

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Explanation:

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3 years ago
The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th
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Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg

where m_a is the mass of the arrow, m_b is the mass of the block, \Delta H of the change in height of the block after the collision, and v is the velocity of the arrow before it hit the block.

Solving for the velocity v, we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} } $

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m_b = 1.40kg,

\Delta H = 0.4m,

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and simplify to get:

\boxed{ v= 15.9m/s}

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