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kolezko [41]
3 years ago
6

If racing alcohol has a mass density of 790kg/m3, what mass will a 1250-litre tank hold?

Physics
1 answer:
Marina86 [1]3 years ago
3 0
Density = Mass per unit Volume
that is,
D = m / V

Now,
We've been provided with,

Density = 790 kg/m³
Volume = 1250 litres = 1.250 m³

Now,
density = \frac{mass}{volume} \\ 790 = \frac{m}{1.250} \\ 790 \times 1.250 = m \\ m = 790 \times 1.250 \\ m = 790 \times \frac{1250}{1000} \\ m = 790 \times \frac{5}{4} \\ m = 197.5 \times 5 \\ m = \frac{1975}{10} \times 5 \\ m = \frac{1975}{2} = 987.5 \: kg

987.5 kg of mass will a 1250 litres tank will hold.

Please Mark It As Brainliest Answer_
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Read 2 more answers
A)Calculate the effective value of g, the acceleration of gravity, at 7900 m above the Earth's surface.
arsen [322]

The solution for the acceleration of gravity is given as

  • g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$
  • g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$

This is further explained below.

<h3>What is the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.?</h3>

Generally,

Mass of earth $M=5.97 \times 10^{24} \mathrm{~kg}$

Radius of earth $R=6371 \mathrm{~km}$

Gravitational const. $G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$

height $h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$$

R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}

In conclusion, acceleration due to gravity at this point will be

g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}}$\\\\$g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}}$\\\\$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$

for $h_{2}=7900 \mathrm{~km}$

R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}}$\\\\$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$

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