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4 years ago

How many grams is 5kg of tap water?

1 answer:

3 0

When you write "5 kg", the little 'k' in the middle stands for 1,000.
So '5 kg' means '5,000 g'.

It doesn't matter what substance you're working with. It could be
tap-water, gold, air, or mud. 5 kg of it is still 5,000 grams of it.

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The bending of light when it changes media is called

olchik [2.2K]

That's called "refraction".

3 0

3 years ago

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where the units of x are length and the numbers 2.6 and 5.1 have appropriate units so that U(x) has units of energy. What is the

KiRa [710]

Answer:

x = 1.00486 m

Explanation:

The complete question is:

" The potential energy between two atoms in a particular molecule has the form U(x) =(2.6)/x^8 −(5.1)/x^4 where the units of x are length and the num- bers 2.6 and 5.1 have appropriate units so that U(x) has units of energy. What is the equilibrium separation of the atoms (that is the distance at which the force between the atoms is zero)? "

Solution:

- The correlation between force F and energy U is given as:

F = - dU / dx

F = - d[(2.6)/x^8 −(5.1)/x^4] / dx

F = 20.8 / x^9 - 20.4 / x^5

- The equilibrium separation distance between atoms is given when Force F is zero:

0 = 20.8 / x^9 - 20.4 / x^5

0 = 20.8 - 20.4*x^4

x^4 = 20.8/20.4

x = ( 20.8/20.4 )^0.25

x = 1.00486 m

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3 years ago

Both the Moon and the Sun influence the tides on Earth. The moon has a much greater influence though. Why is that? A) because th

Temka [501]

C. The sun is 400 times farther from Earth than the moon is.

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3 years ago

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About 65 million years ago an asteroid struck Earth in the area of the Yucatán Peninsula and wiped out the dinosaurs and many ot

9966 [12]

Answer:

$2.44156\times 10^{13}\ m^3$

29010.53917 m

Explanation:

$\rho$ = Density of asteroid = 2 g/cm³

V = Volume

d = Diameter = 10 km

r = Radius = $\dfrac{d}{2}=\dfrac{10}{2}=5\ km$

v = Velocity = 11 km/s

$H_v$ = Heat vaporization of water = $2.26\times 10^6\ J/kg$

$\Delta T$ = Change in temperature = 100-20

Mass is given by

$m=\rho V\\\Rightarrow m=\rho\dfrac{4}{3}\pi r^3\\\Rightarrow m=2000\dfrac{4}{3}\times \pi\times 5000^3\\\Rightarrow m=1.0472\times 10^{15}\ kg$

The kinetic energy is

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1.0472\times 10^{15}\times 11000^2\\\Rightarrow K=6.33556\times 10^{22}\ J$

Heat is given by

$Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg$