Suppose that the Hubble Space Telescope discovers a series of planets with the following characteristics moving around a star th
1 answer:
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Answer:
1317.4 m
Explanation:
We are given that
Angle=
Initial speed =
We have to find the horizontal distance covered by the shell after 5.03 s.
Horizontal component of initial speed=
Vertical component of initial speed=
Time=t=5.03 s
Horizontal distance =
Using the formula
Horizontal distance=
Horizontal distance=1317.4 m
Hence, the horizontal distance covered by the shell=1317.4 m
The Earth's rotational kinetic energy is the kinetic Energy that the Earth
has due to rotation.
The rotational kinetic energy of the Earth is approximately <u>3.331 × 10³⁶ J</u>
Reasons:
<em>The parameters required for the question are; </em>
<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>
<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>
<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>
Which gives;
Therefore;
Which gives;
The rotational kinetic energy of the Earth, = <u>3.331 × 10³⁶ Joules</u>
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<em>The moment of inertia from part A of the question (obtained online) is that of the Earth approximated to a perfect sphere</em>.
<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>
<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>
<em>The rotational period of the Earth, T = 24.0 hrs</em>
Answer:
<em>1.01 W/m</em>
Explanation:
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.
area of the pipe per unit length A = =
m^2/m
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>