Question

solve this plszzzdnd fast ​

Answer 1

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Initial\: velocity\:(u) = 0 \:m/s }$

$\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \: (v) = 2.8 \ m/s }$

$\:\:\:\:\bullet\:\:\:\sf{Acceleration \: (a) = 1.5 \ m/s^{2}}$

${\mathfrak{\underline{\orange{\:\:\:To \:Find:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Time \ taken \ (t) = \:? }$

${\mathfrak{\underline{\red{\:\:\: Solution:-\:\:\:}}}}$

⬤ Using 1st equation of motion :-

$\dashrightarrow\:\: \sf{ v = u + at}$

$\dashrightarrow\:\: \sf{ 2.8 = 0 + 1.5t}$

$\dashrightarrow\:\: \sf{ 2.8 = 1.5t}$

$\dashrightarrow\:\: \sf{ t = \dfrac{2.8}{1.5}}$

$\dashrightarrow\:\: \sf{t = 1.8\overline{6}}$

$\dashrightarrow\:\: \sf{ t = 2 \:(approximately)}$

Hence,

Time taken to reach velocity of 2.8 m/s is $\bf{2\: seconds}$

${\mathfrak{\underline{\pink{\:\:\:Additional \:Information:-\:\:\:}}}}$

$\star\:\sf{Some\:equation\: of \:motion }$

$\boxed{</p><p></p><p>\begin{minipage}{3 cm} \\</p><p></p><p>\sf{\:\bullet\:\:v = u +at} \\ \\</p><p></p><p>\sf{\:\bullet\:\:s = ut + \frac{1}{2}\:at^{2} }\\ \\</p><p></p><p>\sf{\:\bullet\:\:v^{2} = u^{2} + 2as}\\ \\</p><p>\sf{\:\bullet\:\:s = \dfrac{1}{2} (u + v)t}\\ </p><p></p><p>\end{minipage}</p><p></p><p>}$

$\sf{Where,}$

$\:\:\:\:\bullet\:\:\:\textsf{v = Final velocity}$

$\:\:\:\:\bullet\:\:\:\textsf{u = Initial velocity}$

$\:\:\:\:\bullet\:\:\:\textsf{a = Acceleration}$

$\:\:\:\:\bullet\:\:\:\textsf{s = Distance}$

$\:\:\:\:\bullet\:\:\:\textsf{t = Time taken}$

Answer 2

Answer:

Given :

Area of rhombus = 120 cm²

Length of diagonal = 8 cm

To find :

Length of another diagonal

According to the question,

$\sf{ : \implies Area \: of \: rhombus = \dfrac{1}{2} \times d _{1} \times d_{2} }$

$\sf : \implies{ {120 \: cm}^{2} = \dfrac{1}{2} \times 8 \: cm \times x}$

$\sf : \implies{ {120 \: cm}^{2} \times 2 = 8 \: cm \times x }$

$\sf : \implies{ {240 \: cm}^{2} = 8 \: cm \times x}$

$\sf : \implies{ \dfrac{240}{8} \: cm = x}$

${ \underline{ \boxed{ \sf \pink{ : \implies{ \bm3 \bm0 \: c m =x}}}}}$

${ \therefore{ \underline{\sf{So \:,the \: length \: of \: other \: diagonal \: is \: 3 0 \: cm}}}}$