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3 years ago

How many moles of oxygen atoms are in 2 moles of water?

1 answer:

6 0

2 water molecules contain 4 hydrogen atoms and 2 oxygen atoms. a mile of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms !

hope this helped :)

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A (11.4 A) ohm device is connected to a 115 V outlet. If the cost of the electrical energy is $0.16/kWh, determine the cost of r

Scrat [10]

Answer:

$6.68

Explanation:

The current running through the device would be its outlet voltage divided by the resistance:

I = U/R = 115 / 11.4 = 10.09 A

The power generated by this device is the product of its current and voltage

P = IU = 10.09 * 115 = 1160 W or 1.16 kW

If running this device for 36 hours then the total energy it would consume is

E = Pt = 1.16 * 36 = 41.76 kWh

Therefore the total cost of electrical energy is

C = Ec = 41.76*0.16 = $6.68

4 0

3 years ago

A 139 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 89 m below

Illusion [34]

In order to lift the fat (306 lb) physics professor 89 meters up to
the rim, he'll need more potential energy, equal to

(mass) x (gravity) x (height) = (139 x 9.8 x 89) = 121,236 joules .

If the faithful horse delivers 1 constant horsepower = 746 watts,
AND if the cute-as-a-button student has instantly figured out a
way to keep the rope sliding around the edge without any friction,
then the soonest Prof. Tubby can arrive at the rim is

(121,236 joules) / (746 joules/sec) = 162.5 seconds .

Nowhere in this tense drama has the student needed her linguistics
skill yet, but I'll bet it comes in handy as she attempts gamely to
comprehend all of the various pleadings, prayers, and expletives
uttered by her heavy hero from the time he falls over the rim until
he's again lifted to it.

4 0

3 years ago

An astronaut landed on a far away planet that has a sea of water. To determine the gravitational acceleration on the planet's su

TiliK225 [7]

The concept required to solve this problem is hydrostatic pressure. From the theory and assuming that the density of water on that planet is equal to that of the earth $(1000kg / m ^ 3)$ we can mathematically define the pressure as

$P = \rho g h$

Where,

$\rho$ = Density

h = Height

g = Gravitational acceleration

Rearranging the equation based on gravity

$g = \frac{P_h}{\rho h}$

The mathematical problem gives us values such as:

$P = 2.4 atm (\frac{101325Pa}{1atm}) = 243180Pa$

$\rho = 1000kg/m^3$

$h = 28.6m$

Replacing we have,

$g = \frac{243180}{(1000)(28.6)}$

$g = 8.5m/s^2$

Therefore the gravitational acceleration on the planet's surface is $8.5m/s^2$

3 0

2 years ago

A rocket accelerates straight up from the ground at 12.6 m/s^2 for 11.0 s. Then the engine cuts off and the rocket enters free f

FrozenT [24]

Answer:

a) 138.6 m/s

b) 762.3 m

c) 122.3 m/s

d) 24.47

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v=u+at\\\Rightarrow v=0+12.6\times 11\\\Rightarrow v=138.6 \ m/s$

Velocity at the end of its upward acceleration is 138.6 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 12.6\times 11^2\\\Rightarrow s=762.3\ m$

Maximum height the rocket reaches is 762.3 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 762.3-0^2}\\\Rightarrow v=122.3\ m/s$

The velocity with which the rocket crashes to the Earth is 122.3 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 762.3=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{762.3\times 2}{9.81}}\\\Rightarrow t=12.47\ s$