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Ulleksa [173]
3 years ago
11

A ball travels with velocity given by Error converting from MathML to accessible text., with wind blowing in the direction given

by Error converting from MathML to accessible text. with respect to some co-ordinate axes. What is the magnitude of the velocity of the ball in the direction of the wind

Physics
1 answer:
KengaRu [80]3 years ago
5 0

Answer: The magnitude of the velocity = 2/5 m/s

Explanation:

 In this question, the magnitude of the velocity is the product of the magnitude of the displacement vector and the magnitude of the component of the velocity that acts in the direction of displacement.

This will be a scalar projection of V onto X

Please find the attached files for the solution

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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

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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

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3 years ago
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