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2 years ago

A robin in flight has 20.8 J of PE when it is 27.6 m high. What is the mass of the robin? (Unit = kg)

Physics

2 answers:

schepotkina [342]2 years ago

7 0

Answer:

<h2> $\boxed{ \bold{ \purple{0.0769 \: kg \: }}}$ </h2>

Explanation:

$\sf{Potential Energy ( P.E ) \: = \: 20.8 \: joule}$

$\sf{distance \: = 27.6 \: metre}$

$\sf{mass = }$ ?

$\sf{acceleration \: due \: to \: gravity = 9.68 \: {metre \: per \: second}^{2} }$

Now, let's find the mass:

$\sf{PE \: = mass \times gravity \: \times \: distance}$

plug the values

⇒ $\sf{20.8 = m \times 9.8 \times 27.6}$

Multiply the numbers

⇒ $\sf{20.8 = 270.48 \: m}$

Swap the sides of the equation

⇒ $\sf{270.48m = 20.8}$

Divide both sides of the equation by 270.48

⇒ $\sf{ \frac{270.48m}{270.48} = \frac{20.8}{270.48} }$

Calculate

⇒ $\sf{0.0769}$ kg

Hope I helped!

Best regards!!

lora16 [44]2 years ago

6 0

Answer:

$\huge \boxed{\mathrm{0.0769 \ kg}}$

Explanation:

$\displaystyle \sf Mass=\frac{Potential \ energy}{Acceleration \ of \ gravity \times Height}$

$\displaystyle m=\frac{PE}{gh}$

$\mathrm{We \ need \ to \ find \ the \ mass.}$

$\displaystyle m=\frac{20.8}{9.8 \times 27.6}$

$m= 0.0769$

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$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

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Given that $R_1=R_2$ and $l_1=l_2$

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$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

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$\sqrt{3} :\sqrt{10}$

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