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likoan [24]
3 years ago
10

A flagpole is 6.8m tall. A cable is attached from ground to the top of the pole such that it makes a 66 degree angle with the po

le. How long is the cable? How far from the pole is the cable attached to the ground? If the cable is attached 4.2m closer to the pole, what is the new angle between pole and cable?
Physics
1 answer:
soldier1979 [14.2K]3 years ago
3 0

Answer:

Length of cable = 16.7 m

Distance from ground = 15.4

New angle between pole and cable = 58.5 degrees

Explanation:

The flagpole forms a right-angled triangle with the cable and the ground.

Let the adjacent side be the flagpole, the hypotenuse be the cable and the ground be the opposite side.

hypotenuse = h

adjacent = 6.8 m

distance from ground = d

angle = 66.0

cos 66 = 6.8/h

h = 6.8/cos 66

h = 16.7 m

tan 66 = d /6.8

d = 6.8 * tan 66

d = 15.3 m

If the cable is brought 4.2 m closer to the pole, new distance = 15.3 - 4.2 = 11.1 m

let new angle be x

tan x = 11.1/6.8

x = tan⁻¹ (11.1/6.8)

x = 58.5 degrees

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Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

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   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

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          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

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=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

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