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beks73 [17]
3 years ago
11

Find the answer for number 25.

Physics
1 answer:
nirvana33 [79]3 years ago
8 0

<u>We are given:</u>

Mass of load = 140 kg

Acceleration = 0.4 m/s² downward (from Q24)

<u>24. Finding the tension in the cable:</u>

Ma = Mg - T

where 'a' is the acceleration of the load and T is the tension in the cable

replacing the variables, we get:

(140)(0.4)  = (140)(9.8) - T

T = (140)(9.8) - (140)(0.4)

T = 1372 - 56

T = 1316N

<u>25. Finding the distance covered when the load accelerates for 20 s:</u>

in this case, the load will start from rest

so, initial velocity = 0 m/s

acceleration = 0.4 m/s²

from the second equation of motion:

s = ut + (1/2)at²

where s is the distance covered, u is the initial velocity and t is the time taken

s = (0)(20) + (1/2)(0.4)(20)²

s = 0 + 0.2*(400)

s = 80 m

here we didn't take the acceleration due to gravity of the mass into consideration because it is connected to the crane, which would act against any forces and since we are given that the total acceleration is 0.4 m/s², we will just assume that it's the final acceleration of the mass

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 t = 23.255 s,   x = 2298.98 m,    v_y = - 227.90 m / s

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After reading your extensive writing, we are going to solve the approach.

The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.

As there is a mixture of units in different systems we are going to reduce everything to the SI system.

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       0 = y₀ + 0 - ½ g t²

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at the point and arrival the speed is

        vₓ = v₀ₓ = 111.76

     

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         v_y = v_{oy} - gt

          v_y = 0 - gt

          v_y = - 9.8 23.25 555

         v_y = - 227.90 m / s

the negative sign indicates that the speed is down

in the attachment we have a diagram of the movement

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