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zloy xaker [14]
3 years ago
5

Eddie the eagle, british olympic ski jumper, is attempting his most mediocre jump yet. after leaving the end of the ski ramp, he

lands downhill at a point that is displaced 72.1 m horizontally from the edge of the ramp. his velocity just before landing is 33.0 m/s and points in a direction 30.0$^\circ$ below the horizontal. neglect any effects due to air resistance or lift. what was the magnitude of eddie's initial velocity as he left the ramp?
Physics
1 answer:
DedPeter [7]3 years ago
5 0
When he lands his horizontal velocity is 
<span>28 cos40 = 21.45 m/s </span>

<span>the time in flight comes from </span>
<span>x = Vht </span>
<span>58.8 = 21.45t </span>
<span>t = 2.74 seconds </span>

<span>his vertical velocity at landing is </span>
<span>28 sin40 = -18 m/s </span>

<span>his vertical velocity equation is </span>
<span>v = V0 - gt </span>
<span>-18 = V0 - 9.81(2.74) </span>
<span>V0 = -18 + 9.81(2.74) </span>
<span>V0 = 8.88 </span>

<span>his velocity magnitude was </span>
<span>v = (8.88^2 + 21.45^2)^½ </span>
<span>v = 23.2 m/s ANSWER </span>

<span>his initial direction was </span>
<span>tanθ = 8.88/21.45 </span>
<span>θ = 22.5 degrees above the horizontal ANSWER </span>

<span>to find the time to the flight apex from launch </span>
<span>v = gt </span>
<span>8.88 = 9.81t </span>
<span>t = 0.905 s </span>

<span>in 0.905 s Eddie has risen how far above the edge </span>
<span>y = ½(9.81)(0.905^2) </span>
<span>y = 4 m </span>

<span>the remainder of the flight is all drop and takes 2.74 - 0.905 = 1.85 seconds </span>

<span>in 1.85 seconds he drops </span>
<span>y = ½(9.81)(1.85^2) </span>
<span>y = 16.7 m </span>

<span>so the height from the edge to the landing point is </span>
<span>16.7 - 4 = 12.7 m ANSWER</span>
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The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F

<h3>How to determine the coefficient of linear expansion</h3>

From the question given above, the following data were obtained

  • Original diameter (L₁) = 10 m
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The coefficient of linear expansion can be obtained as illustrated below:

α = ∆L / L₁∆T

α = 0.015 / (10 × 90)

α = 0.015 / 900

α = 1.67×10¯⁵ /°F

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A cross-country skier slides horizontally along the snow and comes to rest after sliding a distance of 11 m. If the coefficient
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Answer:

v_o = 4.54 m/s  

Explanation:

<u>Knowns  </u>

From equation, the work done on an object by a constant force F is given by:  

W = (F cos Ф)S                                   (1)  

Where S is the displacement and Ф is the angle between the force and the displacement.  

From equation, the kinetic energy of an object of mass m moving with velocity v is given by:  

K.E=1/2m*v^2                                       (2)

From The work- energy theorem , the net work done W on an object equals the difference between the initial and the find kinetic energy of that object:  

W = K.E_f-K.E_o                                 (3)

<u>Given </u>

The displacement that the sled undergoes before coming to rest is s = 11.0 m and the coefficient of the kinetic friction between the sled and the snow is μ_k = 0.020  

<u>Calculations</u>

We know that the kinetic friction force is given by:

f_k=μ_k*N

And we can get the normal force N by applying Newton's second law to the sled along the vertical direction, where there is no acceleration along this direction, so we get:  

∑F_y=N-mg

     N=mg

Thus, the kinetic friction force is:  

f_k = μ_k*N  

Since the friction force is always acting in the opposite direction to the motion, the angle between the force and the displacement is Ф = 180°.  

Now, we substitute f_k and Ф into equation (1), so we get the work done by the friction force:  

W_f=(f_k*cos(180) s

      =-μ_k*mg*s

Since the sled eventually comes to rest, K.E_f= 0 So, from equation (3), the net work done on the sled is:  

W= -K.E_o    

Since the kinetic friction force is the only force acting on the sled, so the net work on the sled is that of the kinetic friction force  

W_f= -K.E_o  

From equation (2), the work done by the friction force in terms of the initial speed is:  

W_f=-1/2m*v^2  

Now, we substitute for W_f= -μ_k*mg*s, and solving for v_o so we get:  

-μ_k*mg*s = -1/2m*v^2  

v_o = √ 2μ_kg*s

Finally, we plug our values for s and μ_k, so we get:  

v_o = √2 x (0.020) x (9.8 m/s^2) x (11.0 m) = 4.54 m/s  

v_o = 4.54 m/s  

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(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.

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Now, how many milimeter are those?
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