Why are quartz useful

What happens to the radiation coming from the Sun and heading towards Earth?

miss Akunina [59]

Answer: some of the energy are shielded away by the ozone layer,

The rest warm the earth

Explanation:

Not all energy from the sun reaches the earth, some of the energy are shielded away by the ozone layer while the rest energy warm the earth

7 0

3 years ago

In Thomson experiment, why was the glowing beam repelled by a negatively charged plate

algol13

<span>In Thomson experiment, why was the glowing beam repelled by a negatively charged plate, because the glowing beam was negatively charged. The glowing beam particles were attracted to the positive plate.

</span><span>J.JThomson proved that the cathode rays produced a stream of negatively charged particles called electrons. </span>

6 0

3 years ago

Read 2 more answers

Of all the hydrogen in the oceans, 0.0300 % of the mass is deuterium. The oceans have a volume of 317 million mi³.(a) If nuclear

jenyasd209 [6]

Of all the hydrogen in the oceans, 0.0300 % of the mass is deuterium. The oceans have a volume of 319 million mi³. If nuclear fusion were controlled and all the deuterium in the oceans were fused to ⁴₂He, the joules of energy released is E = 6.912× $10^-^1^2$ J

<h3>How is the energy in joules calculated for a controlled nuclear fusion with all the deuterium in oceans fused to ⁴₂He?</h3>

Given the percentage of hydrogen in the ocean = 0.300%

Volume of the ocean = 319 million mi³

Total volume of the ocean, V = 319 × $10^6$ mi³

V = 1.330× $10^1^8$ $m^3$

Density of water = 1000 kg/ $m^3$

Avogadro's number =6.022 × $10^2^3$

There are two hydrogen atoms in a water molecule

Molar mass of Hydrogen =2.016× $10^-^3$ kg/mole

Molar mass of deuterium molecule = 4.028204 × $10^-^3$ kg/mole

Total mass of sea = Volume × Density

= 1.330 × $10^2^1$ kg

Total number of water molecules available in the given mass of water =

[Mass of sea water / Molar mass of water] × Avogadro's number

= 4.446 × $10^4^6$

Total mass of hydrogen in the given mass of water =

[Number of water molecules × Molar mass of hydrogen molecule] / Avogadro's number

= 1.488 × $10^2^0$ kg

Total mass of deuterium in the calculated mass of hydrogen =

Mass of hydrogen × [ density / 100] =

4.464 × $10^1^6$ kg

Total number of deuterium atoms available=

[Total mass of deuterium / Molar mass of deuterium molecule] × Avogadro's number × 2 = 1.335 × $10^4^6$

The combined process consumes 6 deuterium atoms and produces two helium atoms and a total of

E = 43.2× $10^6$ eV of energy.

Therefore the energy release in the consumption of 6 atoms

E=43.2× $10^6$ ×1.6× $10^1^9$ $10^-^1^9$

E=6.912× $10^-^1^2$ J

To learn more about nuclear fusion refer

brainly.com/question/23765509

#SPJ4

4 0

2 years ago

A particle (m = 3.8 × 10-28 kg) starting from rest, experiences an acceleration of 2.4 × 107 m/s2 for 5.0 s. What is its de Brog

Mazyrski [523]

Answer: The De broglie's wavelength is 1.45×10^-14m

Explanation:

Using De broglie's equation

λ= h/mv

Where h= Planck's constant = 6.6×10^-34Js

m= mass in kilogram = 3.8×10^-28kg

V= velocity = acceleration ×time =

2.4×10^7 ×5

V= 1.2×10^8m/s

λ= is the De broglie's wavelength

λ= 6.6×10^-34Js /(3.8×10^-28kg ×1.2×10^8)

λ= 1.45×10^-14m

Therefore ,De broglie's wavelength=

1.45×10^-14m

4 0

4 years ago

A 22.0 kg child plays on a swing having support ropes that are 1.90 m long. A friend pulls her back until the ropes are 45.0 ∘ f

love history [14]

Answer:

120 J

Explanation:

Given:

Mass of the child (m) = 22.0 kg

Length of the rope (L) = 1.90 m

Angle made by the rope (x) = 45.0°

Consider the diagram below for the given scenario.

Point A represents the bottom position, point B represents the maximum height reached, OA or OB is the length of rope.

OA = OB = L = 1.90 m

∠ AOB = x = 45.0°

Now, the maximum height reached by the child is given as:

$h=L-L\cos x\\\\h=L(1-\cos x)\\\\h=1.90\ m(1-\cos (45.0))\\\\h=0.556\ m$

So, as the child is pulled to a height 'h', the child's potential energy is increased.

Therefore, the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing is equal to the increase in the gravitational potential energy of the child.

The increase in gravitational potential energy is given as:

$\Delta U=mgh\\\\\Delta U=(22.0\ kg)(9.8\ m/s^2)(0.556\ m)\\\\\Delta U=119.9\approx 120\ J$

Therefore, the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing is 120 J.

5 0

3 years ago