By using , we can calculate the gravitational attraction as if the solid sphere were a point mass, with all the mass concentrated at the center of the sphere....
1 Answers 1 viewsThe radius= 1.08cm = 0.43in $density=(mass)/(volume)$ $Volume = (mass)/(density)$ $Volume=(82)/(2.7)=30.37cm^3$ The volume of a sphere is given by: $V =4/3pir^3$ So $ r^3=(3V)/(4pi) =(3xx30.37)/(4xx3.142)=1.274$ $r = 1.084cm$
1 Answers 1 viewsFrom definition, is the mass per unit volume. $therefore rho=m/V=m/(4/3pir^3)$ $=(1794g)/(4/3pixx4^3)$ $=6.692g//cm^3$.
1 Answers 1 viewsUse dimensional analysis. $"1 g = 0.00220462 lb"$ $"1 cm"^3=3.53147xx10^(-5) "ft"^3"$ $(1.00cancel"g")/(1cancel"cm"^3)xx(1cancel"cm"^3)/(3.53147xx10^(-5)"ft"^3)xx(0.00220462"lb")/(1cancel"g")="62.7 lb/ft"^3"$
1 Answers 1 viewsLet's break the answer down into three parts: $color (brown) ("First,"$ we're going to convert grams to pounds using this conversion factor: $color(white)(aaaaaaaaaaaaaaaaa$ 1 pound = 453.592 grams $color (purple)...
1 Answers 1 viewsYour goal here will be to use two conversion factors, one to take you from grams to kilograms and one to take you from cubic centimeters to cubic meters....
1 Answers 1 viewsUse "Dimensional Analysis" to determine what "conversion factors" you will need before you do the math. $g/"mL" xx "??"/"??" = "lbs"/"gal"$ We need a mass conversion unit ($g to "lbs"$)...
1 Answers 1 views$rho=2700*kg*m^-3$ $=2700xx10^3*g*(100*cm)^-3$ $=2.700xx10^6*gxx1/(100^3*cm^3)$ $=2.700xx10^6*gxx10^-6*cm^-3$ $=2.700*g*cm^-3$ I always have to double check when I use $"cubic metres"$. A $m^3$ ($-=1000*L$) is a HUGE volume.
1 Answers 1 viewsYour tool of choice here will be the following equation $"% error" = (|"accepted value " - " measured value"|)/"accepted value" * 100%$ So what you're looking for...
1 Answers 1 viewsThe gas will have the same at 546 K and 380 torr that it had at STP. There are two ways to prove this - a longer one and...
1 Answers 1 views