plz do help it's very difficult question!
Answers
0.32% Fe means
0.32g Fe in 100g myoglobin
moles of Fe = 0.32/56 = 0.0057 moles Fe
moles of myoglobin = 100/molar mass = 0.0057 moles myoglobin [as one Fe atom in one myoglobin]
100/MM = 0.0057
100 = 0.57 x MM
100/0.57 = MM
MM = 17500g/mol = 1.75 x 10^4 g/mol
Answer:
1........
Answer :
A body is taken 32 km above the surface of the earth
Radius of earth = 6400 km
Percentage decrease in weight of the body = ?
\quad ━━━━━━━━━━━━━━━━━━
\begin{gathered}\sf :\implies g' = g\bigg\lgroup 1 + \dfrac{h}{R}\bigg\rgroup{}^{-2}\;\; -eq(1)\\\end{gathered}
:⟹g
′
=g
⎩
⎪
⎪
⎪
⎧
1+
R
h
⎭
⎪
⎪
⎪
⎫
−2
−eq(1)
\qquad\quad\dag\:\small\sf h < < R†h<<R
\qquad\quad\dag\:\small\sf \dfrac{h}{R} < < 1†
R
h
<<1
\begin{gathered}\sf :\implies (1+x)^{-2} = 1 - 2x \;\; x < < 1\\\end{gathered}
:⟹(1+x)
−2
=1−2xx<<1
\sf :\implies \bigg\lgroup 1 + \dfrac{h}{R}\bigg\rgroup{}^{-2} = 1 - \dfrac{2h}{R}:⟹
⎩
⎪
⎪
⎪
⎧
1+
R
h
⎭
⎪
⎪
⎪
⎫
−2
=1−
R
2h
\begin{gathered}\sf :\implies g' = g\bigg\lgroup 1 - \dfrac{2h}{R}\bigg\rgroup\;\; -eq(2)\\\end{gathered}
:⟹g
′
=g
⎩
⎪
⎪
⎪
⎧
1−
R
2h
⎭
⎪
⎪
⎪
⎫
−eq(2)
So then the % decrease in the value of acceleration due to gravity is gonna be,
\begin{gathered}\sf :\implies \delta g = \dfrac{g' - g}{g} \times 100\\\end{gathered}
:⟹δg=
g
g
′
−g
×100
\begin{gathered}\sf :\implies \delta g = \dfrac{g\bigg\lgroup 1 - \dfrac{2h}{R}\bigg\rgroup - g}{g} \times 100\\\end{gathered}
:⟹δg=
g
g
⎩
⎪
⎪
⎪
⎧
1−
R
2h
⎭
⎪
⎪
⎪
⎫
−g
×100
\begin{gathered}\sf :\implies \delta g = \dfrac{g\bigg\{ \bigg\lgroup 1 - \dfrac{2h}{R}\bigg\rgroup - 1 \bigg\} }{g} \times 100\\\end{gathered}
:⟹δg=
g
g{
⎩
⎪
⎪
⎪
⎧
1−
R
2h
⎭
⎪
⎪
⎪
⎫
−1}
×100
\begin{gathered}\sf :\implies \delta g = \bigg\{ 1 - \dfrac{2h}{R} - 1 \bigg\} 100\\\end{gathered}
:⟹δg={1−
R
2h
−1}100
\sf :\implies \delta g = - \dfrac{200h}{R} \;\; -eq(3):⟹δg=−
R
200h
−eq(3)
Finally, the % change in weight will be given by,
\begin{gathered}\sf :\implies \delta W = \delta m + \delta g\\\end{gathered}
:⟹δW=δm+δg
\begin{gathered}\sf :\implies \delta W = - \dfrac{200h}{R}\\\end{gathered}
:⟹δW=−
R
200h
\displaystyle \underline{\bigstar\:\textsf{According to the Question :}}
★According to the Question :
\sf\dashrightarrow \delta W = \dfrac{200\times 32}{6400}⇢δW=
6400
200×32
\underline{\boxed{\pink{\mathfrak {\delta w = -1 \%}}}}
δw=−1%