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mezya [45]
3 years ago
7

Why is it inaccurate to use mgy to calculate the potential energy of a satellite orbiting earth at a height one earth radius abo

ve the earth's surface
Physics
1 answer:
lubasha [3.4K]3 years ago
7 0
The equation for the force of gravity is F=Gm(planet)m(satellite)/radius(of orbit)^2


Therefore, the square of the distance between the objects (the earth and the satellite) inversely affects the force of gravity on the satellite. This means that the farther apart the objects are, the smaller the force of gravity is (not the constant acceleration of 9.8m/s^2).
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Which of these rotational quantities is analogous to mass in a linear system?
Rom4ik [11]

Answer:

d

Explanation:

i think it is rotational inertia

because analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in transnational motion.  

hope it's right & helps !!!!!!!!!

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Overcoming an object inertia always requires an
Zinaida [17]
Opposite force in the opposite.
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If two organisms evolve in response to each other, which evolutionary pattern is demonstrated
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3 years ago
Read 2 more answers
Hello!
34kurt
Iodine-131 has a half life of 8 days, so half of it is gone every 8 days.
10 grams of iodine-131 is left for 24 days.
At 8 days: 10/2=5 grams left
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Your mistake is that you stopped at 16 days.
8 0
3 years ago
SHOW WORK
Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\

For Point b:

E=\frac{1}{2} m a^2 w^2\\\\

   =\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J

For Point C:

V_{max}= a w

        = (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}

For point D:

X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm

5 0
3 years ago
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