Wiki User
∙ 7y agoWant this question answered?
Be notified when an answer is posted
An aluminum bar is 2 m long at a temperature of 20° C. What will be the length of the bar at a temperature of 1000 C
r this question....
100 grams isn't really heavy, in fact many chocolate bars weigh in at 100 grams. So anything that has a greater mass than a chocolate bar should also have a mass greater than 100 grams So, a can of soda, a large sandwich, a book.
caliper. Micrometer
A bar graph is for measuring discontinuous data, for example 'What colour eyes do people in this room have'? This would be a bar chart (number on the y axis and eye colours on the x)A line graph is for measuring continuous data, like temperature, for example.
It depends on the density of the material of which the bar is made.
An aluminum bar is 2 m long at a temperature of 20° C. What will be the length of the bar at a temperature of 1000 C
Yes 1 bar = 100 kilopascal = 730,062 torr = 1,01972 kgf/cm2
There are several websites that sell aluminum bar grating. Local hardware stores may also sell aluminum bar grating.
The mass of a bar of soap can vary depending on its size and brand. On average, a standard bar of soap usually weighs between 100-200 grams.
Density = mass / volume Density of aluminum = 34.6 g / 12.8 mL Density of aluminum = 2.70 g/mL
bar graphs are for measuring points of data.
r this question....
The extruded bar would be produced by pushing molten aluminum through a die. Where as the other option for making the bar would be to cast it in a mould. The extruded bar would probably have smoother sides and would be available in longer lengths.
To convert aluminum bus bar length to kilograms, you will need to know its density. The density of aluminum is approximately 2.7 g/cm³. Calculate the volume of the bus bar using its length, width, and height dimensions, then convert the volume to kilograms using the density. A simple formula for this conversion is: Mass (kg) = Density (g/cm³) x Volume (cm³) / 1000.
yes that is how the make them.
Aluminum.