Ask question

Ask question

All categories

3 years ago

14

The total amount of fresh water on earth is estimated to be 3.73 x 10^8 km^3. What is this volume in cubic meters? in cubic feet

?

2 answers:

Brilliant_brown [7]3 years ago

6 0

The total amount of fresh water on earth is $3.73 *10^{8} km^3$

Hence the volume of water $v= 3.73*10^{8} km^{3}$

we are asked to convert this into cubic metre and cubic feet.

we know that 1 km = 1000 m.

$1 km^3 = [10^3 m]^3$

$1 km^3 = 10^{9} m^{3}$

we have been given $v = 3.73 *10^{8} km^3$

$v = 3.73 *10^{8} *10^{9} m^{3}$

$v = 3.73*10^{17} m^{3}$ [ANS]

Now we have to convert it into cubic feet

we know that one metre = 3.28083 feet

$1 m^{3} =[3.28083feet]^{3}$

$1 m^{3} =35.3143472 feet^{3}$

Hence $v =3.73*10^{17} m^{3}$

$= 3.73*10^{17} *35.3143472 feet^{3}$

⇒ $v = 131.7225*10^{17} feet^{3}$ [ANS]

Burka [1]3 years ago

5 0

One km^3 is 1,000,000,000 m^3=10^9 m^3 hence 3.73 10^8 km^3 is 3.73 10^17 m^3

One meter is 3.28084 feet hence 1 m^3 is (3.28084)^3 feet

Thus 3.73 10^8 km^3 is 3.73*35.315 10^17 = 132 cubic feet

You might be interested in

Normalize the equations

tatyana61 [14]

Answer:

Solution is in explanation

Explanation:

part a)

For normalization we have

$\int_{0}^{\infty }f(x)dx=1\\\\\therefore \int_{0}^{\infty }ae^{-kx}dx=1\\\\\Rightarrow a\int_{0}^{\infty }e^{-kx}dx=1\\\\\frac{a}{-k}[\frac{1}{e^{kx}}]_{0}^{\infty }=1\\\\\frac{a}{-k}[0-1]=1\\\\\therefore a=k$

Part b)

$\int_{0}^{L }f(x)dx=1\\\\\therefore Re(\int_{0}^{L }ae^{-ikx}dx)=1\\\\\Rightarrow Re(a\int_{0}^{L }e^{-ikx}dx)=1\\\\\therefore Re(\frac{a}{-ik}[\frac{1}{e^{ikx}}]_{0}^{L})=1\\\\\Rightarrow Re(\frac{a}{-ik}(e^{-ikL}-1))=1\\\\\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1$

$\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1\\\\\frac{a}{k}Re(icos(-kL)+sin(kL)+\frac{1}{i})=1\\\\\frac{a}{k}sin(kL)=1\\\\a=\frac{k}{sin(kL)}$

7 0

3 years ago

kykrilka [37]

Answer:

Clouds form when below the dew point

4 0

3 years ago

Ok, so this question is probably really easy but I can't really be bothered to answer it, terrible I know, but I thank all usefu

Dvinal [7]

Without a bulb energy cant go through and it would be an open circuit blocking the energy from coming out.

3 0

3 years ago

Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

$R=\rho \dfrac{l}{A}$

$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

or

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0

2 years ago

Which equation is used to calculate the weight of an object on a planet?

Kamila [148]

In this case to find the weight of an object you must use the formula.

W = mg

6 0

3 years ago

Read 2 more answers