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Naddik [55]
2 years ago
14

The food chain in question 1 is; Grass -> Deer -> Wolf

Physics
1 answer:
aleksklad [387]2 years ago
6 0
Grass dear wolf is the right awnser
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Modern observations have shown that the geometry of the universe is ____.
lesantik [10]

Modern observations have shown that the geometry of the universe is flat and the universe mist be infinite.

<h3>What is geometry?</h3>

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1 year ago
What does the number, or density, of field lines on a source charge indicate?
Harrizon [31]

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6 0
3 years ago
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John ( body mass= 65 kg) is taking off for a long jump . Horizontal accerleration ax is 5m/s^2 and vertical acceleration ay is 0
cestrela7 [59]

Answer:

(a) The horizontal ground reaction force  F_{g,x}=325\, N

(b)  The vertical ground reaction force F_{g,y}=696\, N

(c)   The resultant ground reaction force F_g=768\, N

Explanation:  

Given

John mass , m = 65 kg

Horizontal acceleration , a_x= 5.0 \frac{m}{s^{2}}

Vertical acceleration , a_y=0.9 \frac{m}{s^{2}}

(a) Using Newton's 2nd law in horizontal direction

F_{g,x}=ma_x

=>F_{g,x}=65\times 5\, N=325\, N

Thus the horizontal ground reaction force  F_{g,x}=325\, N

(b) Using Newton's 2nd law in vertical direction

F_{g,y}-mg=ma_y

=>F_{g,y}=mg+ma_y

=>F_{g,y}=65\times (9.81+0.9)\, N=696\, N

Thus the vertical ground reaction force F_{g,y}=696\, N

(c)  Resultant ground reaction force is

F_g=(F_{g,x}^{2}+F_{g,y}^{2})^{\frac{1}{2}}

=>F_g=(325^{2}+696^{2})^{\frac{1}{2}}\, N=768\, N

=>F_g=768\, N

Thus  the resultant ground reaction force F_g=768\, N

5 0
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}}} \\

4 0
2 years ago
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