Question

$calculate \: log_{10}36$
​

Answer 1

Answer:

Solution :

Given log₁₀2 = 0.3010 , log₁₀5 = 0.6990

Now, log₁₀50 - log₁₀40

  = log₁₀(5² × 2) - log₁₀(2³ × 5)

  = (log₁₀5² + log₁₀2) - (log₁₀2³ + log₁₀5)

  = 2log₁₀5 + log₁₀2 - 3log₁₀2 - log₁₀5

  = log₁₀5 - 2log₁₀2

  = 0.6990 - 2(0.3010)

  = 0.6990 - 0.6020

  = 0.0970

➞ log₁₀50 - log₁₀40 = 0.970 , which is the required value correct to 3 significant figures.

Answer 2

Explanation:

Pressure exerted by water on the bottom of a deep dam(Hydrostatic pressure) is 120000 Pa.

Step-by-step Explanation :

GivEn :

Density of water, ϱ = 10³kg/m³

Height = 12 m

g = 10m/s²

To find :

Pressure exerted by water on the bottom of a deep dam(Hydrostatic pressure) =

SoluTion :

We know that,

\begin{gathered}\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = \rho \times g \times h \\ \\\end{gathered}\end{gathered}

⟶Hydrostaticpressure=ρ×g×h

Substituting the values,

\begin{gathered}\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = {10}^{3} \times 10 \times 12 \\ \\\end{gathered}\end{gathered}

⟶Hydrostaticpressure=10

3

×10×12

\begin{gathered}\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = {10}^{4} \times 12 \\ \\\end{gathered}\end{gathered}

⟶Hydrostaticpressure=10

4

×12

\begin{gathered}\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = 10000 \times 12 \\ \\\end{gathered}\end{gathered}

⟶Hydrostaticpressure=10000×12

\begin{gathered}\begin{gathered}\sf \therefore { \blue{Hydrostatic \: pressure = 120000 \: Pa}} \\ \\\end{gathered}\end{gathered}

∴Hydrostaticpressure=120000Pa

Hence, Pressure exerted by water on the bottom of a deep dam(Hydrostatic pressure) is 120000 Pa.