1 g = 1 ÷ 1000 kg
= 0.001 kg
1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³
1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³
The density is 1000 kg/m³.
To find the surface area of a single cube we first nees to take the cube root of 8cm3 which is 2.
Now we know that the length of each side is 2 and we can find the area of one side by doing 2x2 which is 4.
To find the total surface area of one cube we do 4 times 6 side giving us a total of 24cm2.
To find the total surface area of the 8 individual cubes, we multiply 24cm2 by 8 to give us a total of 192cm2.
Now to find the total surface area of the one large cube, we know that each side of one of the small cubes is 4cm2 and the large cube is set up so that there are two levels of four cubes right on top of each other. So, the total area of each side of the large cube is 4cm2 times 4 which gives us 16cm2.
Then we multiply 16cm2 by 6 sides to give us a total surface area of 96cm2.
The ratio of the surface area of the single large cube comapred to the total surface area of the single cubes is 96:192
We can further simplify this ratio:
96:192
48:96
24:48
12:24
6:12
3:6
1:2
Data is inappropriate
here, we need gauge of the wire i.e., diameter of the wire, so that we calculate the resistance by using the formula
R = ρl/A
where R= resistance ; Ω
l = length of wire ; m
A = area of wire ; m²
ρ = resistivity ; Ω-m
But in general ohms law is
V = I R
R = V/I ;
but here we also calculate "R" from length of wire in which the current is flowing.
I hope it is helpful to you.
Answer:Explanation:
Image result for what does a worm and wheel mechanism do to torque and speed
Like other gear arrangements, a worm drive can reduce rotational speed or transmit higher torque. ... Each full 360 degree turn of a single start worm advances the gear by one tooth. For a multi start worm the gear reduction equals the number of teeth on the gear divided by the number of starts on the worm.
Answer:
which of the cars are speeding up: c
which of the cars or slowing down: a
which of the cars are maintaning a constant speed: b
Explanation:
