The distance between the curve and the line where it approaches zero as they tend to infinity is the line in the asymptote of a curve. This is unusual for modern authors but in some sources the requirement that the curve may not cross the line infinitely often is included.

The point that does not fit the rest of the graph or is undefined is called a removable discontinuity. By filling in a single point, the removable discontinuity can be made connected.

6 0

3 years ago

A cylindrical can 150 mm in diameter is filled to a depth of 100 mm with a fuel oil. The oil has a mass of 1.56 kg. Calculate it

Kazeer [188]

Answer:

Density (φ) = 0,8827 Kg/L

Specific weight (Ws) = 8,65 N/L

Specific gravity (Gs) = 0,8827 (without unit)

Explanation:

The density formula: φ = $\frac{m}{V}$

I know the mass "m", I need to find out the volume of the cylinder (V)

V = π* r²*h

The radius "r" is equal to half the diameter (150mm) = 75mm

Now I can find out the density (φ)

φ = $\frac{1,56Kg}{1,767145L}$ = 0,8827 Kg/L

The specific weight (Ws) is the relationship between the weight of substance (oil) and its volume. We apply the following formula:

Ws = φ*g

(g = gravity = 9,8 m/s²)

Finally, specific gravity (Gs) is the ratio between the density of a substance (oil) "φ(o)" and the density of water "φ(w)" :

Gs = φ(o) / φ(w)

(φ(w) = 1 Kg/L

Hope this can help you !!

3 0

3 years ago

The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m

Sonbull [250]

Answer:

density is $10^{6}$ Mg/µL

Explanation:

given data

density of nuclear = $10^{18}$ kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

1 m³ is equal to $10^{6}$ cm³

and here 1 cm³ is equal to 1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = $10^{18}$ kg/m³

density = $10^{18}$ kg/m³ × $\frac{10^{-3} }{10^{6} }$ Mg/µL

density = $10^{6}$ Mg/µL

8 0

3 years ago

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