, so you insert the proportion and get 18. When the tank is three-fifths full, it contains 18 gallons of gas.
6.0⋅<span>102</span><span>kg m<span>−3</span></span>
Explanation:
In order to find the density of the bowling ball, <span>ρ
</span>, you must determine the mass one unit of volume of this bowling ball would have.
Notice that the problem provides you with the volume of the bowling ball expressed cubic meters, <span><span>m3</span>
</span>, which means that one unit of volume would be <span><span>1 m3</span>
</span>.
Now, you know that a volume of <span><span>0.0050 m3</span>
</span> has a mass of <span>3.0 kg
</span>. You can use this known proportion as a conversion factor to figure the mass of <span><span>1 m3</span>
</span>
<span><span>1<span>m3</span>⋅<span>3.0 kg<span>0.0050<span>m3</span></span></span>=600 kg</span>
</span>
So, you know that one unit of volume has a mass of <span>600 kg
</span>, which means that the bowling ball's density is equal to
<span><span><span>ρ=6.0⋅<span>102</span><span>kg m<span>−3</span></span></span><span>−−−−−−−−−−−−−−−−−</span></span>
</span>
The answer is rounded to two sig figs.
Undefined
If it’s wrong Oml
It would be the side-angle-side theorem
The equation is A
explanation:
