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Paul [167]
3 years ago
12

HELP NOW PLS !!

Physics
1 answer:
Kitty [74]3 years ago
7 0

Answer:

The value is s =197.88 \  m  

Explanation:

From the question we are told that

    The  impulse is  I =  950 \  N\cdot s

    The mass is  m  =  12 \  kg

     The time t =  5 s

Generally impulse is mathematically represented as

         I =  F *  t

=>       F =  \frac{I}{t}

=>       F =  \frac{950 }{ 5 }

=>       F = 190 \  N

Generally force is mathematically represented as

      F =  m * a

=>    a =   \frac{F }{ m }

=>     a =   \frac{190 }{ 12  }

=>     a =   15.83 \  m/s^2

Generally from kinematic equation , the distance covered is  

      s =  ut +  \frac{1}{2}  * at^2

Here  u is the initial velocity of the cart and the value is u =  m/s

=>   s =  0 * 5 +  \frac{1}{2}  *  15.83 * 5^2  

=>   s =197.88 \  m  

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notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

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V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

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Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

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