The mass of the earth is 5.97x10 24 kilograms. the mass of the moon is 7.34 x 10 22 kilograms

1 answer:

3 0

7.34x10^22 kg can also be written as 0.0734 * 10^24 kg

Since both numbers are * 10^24, you can just add 5.97 + 0.0734 = 6.043. Then you have to add the * 10^24 giving you 6.043 * 10^24 kg

The ratio of Earth / moon is 5.97x10^24 / 7.34x10^22 = 81. The mass of the Earth is 81 times the mass of the moon.

You might be interested in

How do you simplify 4c=-r c=-9 and r=-8

frutty [35]

Answer: -36 = 8

Step-by-step explanation:

$4c=-r\\4(-9)=-(-8)\\-36=8$

3 0

2 years ago

Read 2 more answers

A metalworker has a metal alloy that is 15% copper and another alloy that is 75% copper. How many kilograms of each alloy should

attashe74 [19]

X - the mass of the first alloy (15%)
y - the mass of the second alloy (75%)

x kg of the first alloy contains 15%x=0.15x=(15/100)x=(3/20)x kg of copper
y kg of the second alloy contains 75%y=0.75y=(75/100)y=(3/4)y kg of copper
90 kg of a 51% copper alloy contains 51%*90=0.51*90=45.9 kg of copper

$x+y=90 \\
\frac{3}{20}x+\frac{3}{4}y=45.9 \ \ \ |\times (-\frac{4}{3}) \\ \\
x+y=90 \\
\underline{-\frac{1}{5}x-y=-61.2} \\
x-\frac{1}{5}x=90-61.2 \\
\frac{4}{5}x=28.8 \ \ \ |\times \frac{5}{4} \\
x=36 \\ \\
36+y=90 \\
y=90-36 \\
y=54$

The metal worker should combine 36 kg of the 15% copper alloy and 54 kg of the 75% copper alloy to create 90 kilograms of a 51% copper alloy.

7 0

2 years ago

The English alphabet has 20 letters more than one fourth the number of letters in the Greek alphabet. The English alphabet has 2

Black_prince [1.1K]

1/4G+20=26
G=24
I cannot really explain this because it is hard to put into words, but it is correct.

7 0

3 years ago

Translate the sentence into an equation

STatiana [176]

Answer:

5-3x=7

Step-by-step explanation:

8 0

2 years ago

Let c be the curve of intersection of the parabolic cylinder x2 = 2y, and the surface 3z = xy. find the exact length of c from t

Mandarinka [93]

Parameterize the intersection by setting $x(t)=t$ , so that

$x^2=2y\iff y=\dfrac{x^2}2\implies y(t)=\dfrac{t^2}2$
$3z=xy\iff z=\dfrac{xy}3\implies z(t)=\dfrac{t^3}6$

The length of the path $C$ is then given by the line integral along $C$ ,

$\displaystyle\int_C\mathrm dS$

where $\mathrm dS=\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dz}{\mathrm dt}\right)^2}\,\mathrm dt$ . We have

$\dfrac{\mathrm dx}{\mathrm dt}=1$
$\dfrac{\mathrm dy}{\mathrm dt}=t$
$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{t^2}2$

and so the line integral is

$\displaystyle\int_{t=0}^{t=2}\sqrt{1^2+t^2+\dfrac{t^4}4}\,\mathrm dt$

This result is fortuitous, since we can write

$1+t^2+\dfrac{t^4}4=\dfrac14(t^4+4t^2+4)=\dfrac{(t^2+2)^2}4=\left(\dfrac{t^2+2}2\right)^2$

and so the integral reduces to

$\displaystyle\int_{t=0}^{t=2}\frac{t^2+2}2\,\mathrm dt=\dfrac{10}3$

3 0

3 years ago