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Eva8 [605]
3 years ago
9

How did Aristotle's inability to detect parallax lead him to propose a geocentric model of the Solar System.

Physics
1 answer:
sasho [114]3 years ago
6 0
He reasoned that since parallax could not be observed for celestial objects near the sun, then the earth was stationary. This erroneous assumption was because at the time he had no way of knowing that celestial objects were so far away that their parallax angles were too small to detect.
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What is the change in velocity of a 22-kg object that experiences a force of 15 N for
vagabundo [1.1K]

Answer:

Force = mass × acceleration

Acceleration:

{ \tt{15 = (22 \times a)}} \\ { \tt{a =  \frac{15}{22}  \:  {ms}^{ - 2} }}

From first Newton's equation of motion:

{ \bf{v = u + at}}

Change = v - u:

{ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \:  {ms}^{ - 2} }}

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3 years ago
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before
Darina [25.2K]

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

\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before  \: it \: comes \: to \: rest \:( s_{2} )}

\\

{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity   = v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \:  s_{1} \: penetration =  \dfrac{v}{2}  \:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{s_{1} =  \dfrac{3}{100}  = 0.03 \: m}

\\

☯ As we know that,

\\

\dashrightarrow\:\: \sf{ {v}^{2}  =  {u}^{2} + 2as }

\\

\dashrightarrow\:\: \sf{  \bigg(\dfrac{v}{2} \bigg)^{2}  =  {v}^{2}   + 2a s_{1}}

\\

\dashrightarrow\:\: \sf{  \dfrac{ {v}^{2} }{4}  =  {v}^{2}  + 2 \times a \times 0.03  }

\\

\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4}  -  {v}^{2}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{\dfrac{ -  3{v}^{2} }{4}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{a =  \dfrac{ - 3 {v}^{2} }{4 \times 0.06}  }

\\

\dashrightarrow\:\: \sf{ a =  \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }

\\

\:\:\:\:\bullet\:\:\:\sf{  Initial\:velocity=v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }

\\

\dashrightarrow\:\: \sf{  {v}^{2}  =  {u}^{2}  + 2as}

\\

\dashrightarrow\:\: \sf{{0}^{2}  =  {v}^{2}  + 2 \times  \dfrac{ - 25 {v}^{2} }{2}  \times s  }

\\

\dashrightarrow\:\: \sf{ -  {v}^{2}  =  - 25 {v}^{2}  \times s  }

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{ -  {v}^{2} }{ - 25 {v}^{2} }}

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{1}{25} }

\\

\dashrightarrow\:\: \sf{ s = 0.04 \: m }

\\

☯ For left penetration (s₂)

\\

\dashrightarrow\:\: \sf{s =  s_{1} +  s_{2}  }

\\

\dashrightarrow\:\: \sf{  0.04 = 0.03 +  s_{2}}

\\

\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }

\\

\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}

\\

\star\:\sf{Left \: penetration \: before  \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\

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Forest fires can be an impact 

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We know, P = F . ΔV

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2 years ago
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