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gregori [183]
3 years ago
13

1. An object is accelerated at a rate of 5 m/s²? Find the mass of the object if the force exerted on it is 4 x 10⁵dynes. (Note:

1 N = 1 10⁵ dynes)
Given:
Unknown:
Formula:
Solution:
Final answer:​​
Mathematics
1 answer:
torisob [31]3 years ago
8 0

Step-by-step explanation:

\begin{gathered}\frak \red{Given}\:\begin{cases} \:\quad\sf  Acceleration \:of \ \: a\:body\:,\:a\:=\:\pmb{\frak{5\:m/s^2\:}}\:\\ \:\quad\sf Force \:Exerted\:on\:a\:body\:,\:F\:=\:\pmb{\frak{\:4\times 10^5\: dynes\:}}\:\end{cases}\\\\\end{gathered}

<h2><u>Need To Find : </u></h2>

  • The mass of a body ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

We're Provided with the , Acceleration (a) , and Force Exerted (F) on a body and , we'll find Mass (m) of the object using the <u>relation</u>

 between <u>Force , Mass & Acceleration </u> and, that is : 

\underline {\pmb{\sf \: F\:=\:m\:\times \:a \:}}

\\ \twoheadrightarrow \sf F \:=\:m\:\times \:a

: \rightarrow \sf 4 \:\times 10^5 \:=\:m\:\times \:5\:

\\  \\  \\ \\\twoheadrightarrow \sf 4 \:=\:m\:\times \:5\:\:\Bigg\lgroup \:1 \:N\:=\:1\:\times 10^5\:dynes\:\Bigg\rgroup\\

\\ \twoheadrightarrow \sf 4 \: \:=\:m\:\times \:5\:\\\\\\

\twoheadrightarrow \sf 5m\:=\:4\:\\

\twoheadrightarrow \sf m \cancel{\dfrac{4}{5}}\\\\

\ \twoheadrightarrow \pmb {\underline {\boxed {\purple {\:\frak{ \:m\:\:=\:0.8\:kg\:}}}}}\:\bigstar \: \\

\begin{gathered}\therefore \:\underline {\sf \: Hence,\:Mass\:of\:a\:body\:is\:\pmb{\sf 0.8\:kg\:}\:.}\:\\\\\end{gathered}

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Step-by-step explanation:

Two lines are given to us which are perpendicular to each other and we need to find out the value of a . The given equations are ,

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Step 1 : <u>Conver</u><u>t</u><u> </u><u>the </u><u>equations</u><u> in</u><u> </u><u>slope</u><u> intercept</u><u> form</u><u> </u><u>of</u><u> the</u><u> line</u><u> </u><u>.</u>

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Step 2: <u>Find </u><u>the</u><u> </u><u>slope</u><u> of</u><u> the</u><u> </u><u>lines </u><u>:</u><u>-</u>

Now we know that the product of slope of two perpendicular lines is -1. Therefore , from Slope Intercept Form of the line we can say that the slope of first line is ,

\sf\longrightarrow Slope_1 = \dfrac{3}{2}

And the slope of the second line is ,

\sf\longrightarrow Slope_2 =\dfrac{-6}{a}

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\sf\longrightarrow \dfrac{3}{2}\times \dfrac{-6}{a}= -1

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\sf\longrightarrow \boxed{\blue{\sf a = 9 }}

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Step-by-step explanation:

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