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leonid [27]
2 years ago
8

An object with velocity 141 ft/s has a kinetic energy of 1558.71 ft∙lbf, on a planet whose gravity is 31.5 ft/s2. What is its

mass in pounds
Physics
1 answer:
Sidana [21]2 years ago
6 0

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m  = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \  \times \ \ \ \ 1   \ lbf\ }

m  = \frac{(2 \ \times \ 1558.71  \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71  \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm

Therefore, the mass of the object is 5.045 lbm.

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Answer:

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Explanation:

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This is a general physical law derived from

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when distance is doubled the gravitational force will be reduced by quarter not half.

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3 years ago
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In 1-2 sentences explain how you would determine the constant k for a spring.
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The weight is now plotted on the vertical axis and the extension is plotted on the horizontal axis. The slope of the graph is the force constant of the spring.

Learn more: brainly.com/question/10991960

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3 years ago
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This is an example of conduction
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The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr
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7 0
3 years ago
Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.0 kg⋅m2 and for arms and legs in is 0.90
babymother [125]

Answer:

Her angular speed (in rev/s) when her arms and one leg open outward is 1.56\frac{rev}{s}

Explanation:

Initial moment of inertia when arms and legs in is I_i=0.90 kg.m^{2}

Final moment of inertia when her arms and on leg open outward, I_f=3.0 kg.m^{2}

Initial angular speed w_i=5.2\frac{rev}{s}

Let the final angular speed be w_f

Since external torque on her is zero so we can apply conservation of angular momentum

\therefore L_f=L_i

=>I_fw_f=I_iw_i

=>w_f=\frac{I_iw_i}{I_f}=\frac{0.9\times5.2 }{3.0}\frac{rev}{s}=1.56\frac{rev}{s}

Thus her angular speed (in rev/s) when her arms and one leg open outward is 1.56\frac{rev}{s}

7 0
3 years ago
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