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statuscvo [17]
2 years ago
12

How much force is needed to keep the 750000 Newton Space Shuttle moving at a constant speed of 28000 km/h, in a straight line?

Physics
1 answer:
Alika [10]2 years ago
6 0

The force needed to keep the space shuttle moving at constant speed is 0.

The given parameters;

  • <em>weight of the space shuttle, F = 750,000 N</em>
  • <em>constant speed of the space shuttle, v = 28,000 km/h</em>

The mass of the space shuttle is calculated as follows;

W = mg\\\\m = \frac{W}{g} \\\\m = \frac{750,000}{9.8} \\\\m = 76,530.61 \ kg

The force needed to keep the space shuttle moving at constant speed is calculated as follows;

F = ma

F = 76,530.61 \times a

where;

a is the acceleration of the space shuttle

At a constant speed, acceleration is zero.

F = 76,530.61 x 0

F = 0

Thus, the force needed to keep the space shuttle moving at constant speed is 0.

Learn more here:brainly.com/question/16374764

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A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o
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Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) =  \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_i^2) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2  ) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 -  \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 -  \frac{3}{4}v_f^2\\\\

h = \frac{3}{4g}(v_1^2 -v_f^2)

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0
3 years ago
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