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A bone has a mass of 5.97 grams and a volume of 3.00 cubic centimeters . Calculate its destiny

Anuta_ua [19.1K]

1.99 is the density because of d= m/v

7 0

3 years ago

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Why does air resistance affect the location of and objects trajectory?

mafiozo [28]

<span>When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. Air resistance depends on two important factors - the speed of the object and its surface area. Increasing the surface area of an object decreases its speed.</span>

3 0

3 years ago

Which of the following is true of physical properties? Physical properties include properties like mass, color, and flammability

nataly862011 [7]

Answer:

B) Physical properties cannot be observed without changing the identity of the material.

6 0

4 years ago

Stars produce their own light but planets do not. Even then we can see them. explain

PilotLPTM [1.2K]

Planets are not hot enough, and that's because they lack the internal energy sources that stars have: thermonuclear fusion (They do emit a little light, mainly infrared light, but nothing compared to the sun)

Planets don't have their own light, so simply they do not emit light, this phenomena goes back to their information. When they are formed, they are much, much less hotter than the stars. In stars, fusion reaction always goes on, in which energy, as well as light is formed, so, there is light for stars to emit for a long time.

4 0

4 years ago

The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune

Veronika [31]

Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
$\frac{r^3}{T^2}= \frac{GM}{4 \pi^2}$ (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
$1 AU = 1.5 \cdot 10^{11} m$
so the radius of the orbit is
$r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m$

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
$T=\sqrt{ \frac{4 \pi^2}{GM} r^3} = \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s$

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
$T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years$

3 0

3 years ago

What belongs in the center section ​

What belongs in the center section