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2 years ago

HELP ASAP. WILL GIVE BRAINLIEST.

1 answer:

3 0

The fastest thing in the whole universe is the speed of light in a vacuum. The speed is around 2.99 * $10^{8}$ . It is fastest in vacuum as vacuum does not contain any dust particles and is a hollow space so light travels faster. It also has a refractive index of 1 so it travels faster.

After vacuum light travels faster in air. Air has a refractive index of 1.0003.

After air, it travels faster in water. The speed is about 3/4 of its speed in a vacuum, or 225,000,000 meters per second. Speed slows in water because photons interact with atoms, which repeatedly absorb and re-emit them to the same direction. The refractive index of water is 1.33.

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What’s the difference between the offspring of an asexual organism and a sexual organism

laiz [17]

The offspring of an asexual organism (since asexual reproduction only requires one parent) will be completely identical to the parent, like bacteria or fungi. The offspring of a sexual organism (since sexual reproduction requires two organisms) will be a combination of both parents, like animals or (some) flowers.

4 0

2 years ago

Read 2 more answers

The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc

Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is $1.163 * 10^{-5} m$

Explanation:

From the uncertainty Principle,

Δ $_{y}$ Δ $_{p}$ $= \frac{h}{2\pi }$

The momentum P $_{y}$ = (mass of electron)(speed of electron)

= $(9.109 * 10^{-31}kg)(995 * 10^{3} m/s)$

= $9.0638 * 10^{-25}kgm/s$

If the uncertainty is reduced to a 0.0010%, then momentum

= $9.068 * 10^{-30}kgm/s$

Thus the uncertainty in the position would be:

Δ $_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }$

Δ $_{y} \geq 1.163 * 10^{-5}m$

5 0

3 years ago

A comet fragment of mass 1.96 × 1013 kg is moving at 6.50 × 104 m/s when it crashes into Callisto, a moon of Jupiter. The mass o

vredina [299]

Answer:

Recoil speed, $1.17\times 10^{-5}\ m/s$

Explanation:

Given that,

Mass of the comet fragment, $m_1=1.96\times 10^{13}\ kg$

Speed of the comet fragment, $v_1=6.5\times 10^4\ m/s$

Mass of Callisto, $m_2=1.08\times 10^{23}\ kg$

The collision is completely inelastic. Assuming for this calculation that Callisto's initial momentum is zero. So,

$m_1v_1=(m_2+m_2)V$

V is recoil speed of Callisto immediately after the collision.

$V=\dfrac{m_1v_1}{(m_2+m_2)}\\\\V=\dfrac{1.96\times 10^{13}\times 6.5\times 10^4}{(1.96\times 10^{13}+1.08\times 10^{23})}\\\\V=1.17\times 10^{-5}\ m/s$