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3 years ago

Which of the following water molecules exists in the greatest abundance?

Physics

1 answer:

aliina [53]3 years ago

7 0

Answer:

d. H216O

Explanation:

Hello,

Water is a chemical compound which has two hydrogen atoms and one oxygen atom.In solid state, it is ice where as in gas state, it is steam.Hydrogen and oxygen atoms in water molecule form polar covalent bonds formed through hydrogen bonding. Oxygen has three isotopes to remember ; H2160, H2180 and H2170. Sea water for example has approximately 99% H2160 and less than 0.5% H2180.

Evaporating waters releases more H2160 to the atmosphere than H2180.In areas where polar ice takes time to melt, the oceans are contantly depleted of H2160 until it melts to return to the water bodies and maintain balance.This shows that the water molecules that exists in greatest abundance is H2160.

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Water drips from a shower head (the sprayer at the top of the shower) and falls onto the floor 2.3 m below. The droplets are fal

Maurinko [17]

Answer:

0.767m

Explanation:

We are given that the time interval between each droplet is equal.

We are also given that the fourth drop is just dripping from the shower when the first hits the floor.

If they fall at the same time interval and we know that the distance between the shower head and floor are the same, they must therefore fall at the same velocity.

The distance between each drop has to be the same given that they fall at equal time intervals.

Let this distance be x.

We can then partition the entire height of the system into three parts (as shown in the diagram).

Hence, we can say that:

x + x + x = 2.3m

3x = 2.3m

=> x = 2.3/3 = 0.767m

Therefore, at the time the first drop hits the floor, the third drop is only 0.767 m below the shower head.

8 0

3 years ago

Which property of a body resists change from a state of rest or of motion?

jeyben [28]

Answer:

mass

Explanation:

As this is directly related to inertia. Inertia is the property of unit which resists the change of state of rest or motion

7 0

2 years ago

Read 2 more answers

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$