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kherson [118]
3 years ago
15

A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her accelerat

ion when her speed is 30.0 m/s
Physics
1 answer:
kirill [66]3 years ago
4 0

Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

784 - 0.3136 * 900 = 80a

784 - 282.24 = 80a

497.76 = 80a

a = 497.76 / 80

a = 6.22 m/s²

Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²

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Answer:

Fe= 28.2 N : Magnitude of the equilibrant (Fe)

β = 18.34° , clockwise from the positive x axis

Explanation:

Concept of the equilibrant

It is called equilibrant  to a force with the same magnitude and direction as the resulting one (in case it is non-zero) but in the opposite direction. Adding vectorially to all the forces (that is to say the resulting one) with the equilibrant you get zero

To solve this problem we decompose the forces given into x-y components to find the resulting force:

Look at the attached graphic

F₁= 33.4 N  , θ₁=23.8° clockwise from the positive y axis (y+)

F₁x= 33.4 *sin23.8° = 13.48 N

F₁y= 33.4 *cos23.8° =30.6 N

F₂=46.1 N ,  θ₂=28.8 counterclockwise from the negative x axis (x-)

F₂x= -46.1 *cos28.8° = -40.4 N

F₂y=  -46.1 *sin28.8° =  -22.2 N

Components of the resultant in x-y R(x,y)

Rx= 13.48 N -40.4 N = - 26.92 N

Ry= 30.6 N  -22.2 N =  + 8.4 N

Components of the equilibrant in x-y Fe(x,y)

Fex= +26.92 N

Fey=  - 8.4 N

Magnitude of the equilibrant (Fe)

F_{e} = \sqrt{(F_{ex})^{2}+{(F_{ey})^{2}  }

F_{e} = \sqrt{(26.92)^{2}+(8.4)^{2}  }

Fe= 28.2 N

Angle the equilibrant makes with the x axis ( β)

\beta = tan^{-1} (\frac{F_{ey} }{F_{ex} } )

\beta = tan^{-1} (\frac-8.4 }{26.92 } )

β = -18.34°                  

β = 18.34° , clockwise from the positive x axis

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KCL at Junction a. </span><span>
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total charge  lying inside is +2q - 2q = 0 . So in the thickness of the shell , electric field will be zero as total charge inside is nil.

For a point at r > b total charge inside is 2q-2q+10q = 10q , so electric field at r which is lying outside  the shell .

E = k 10 q / r² for r > b

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