3 years ago

Look at your data. With three foods, Flock v

Physics

1 answer:

wolverine [178]3 years ago

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Answer:the first one was x

the second one is y

Explanation:

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A. It does not exhibit projectile motion and follows a straight path down the ramp.

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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 $F_{g,y}=696\, N$

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

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

=> $F_g=768\, N$

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A student has a displacement of 739 m north in 162 s. What was the student’s average velocity?A. 0.22 m/sB. 119,718 m/sC. 162 m/

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Answer:

answer below

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Displacement of the student is 739 m due North and it takes 162 s.

We need to find the student's average velocity. Using formula of velocity.

Velocity = displacement/time

v= 739/162

v= 4.56

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