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Ostrovityanka [42]

2 years ago

3

Consider Earth to be stationary, and the moon as orbiting Earth in a circle of radius R. If the masses of Earth and the Moon are

ME and Mm, respectively, which of the following best represents the total mechanical energy of the Earth-Moon system?a. $\frac{GM_EM_M}{2R}$
b. $\frac{GM_EM_M}{R}$
c. $-\frac{GM_EM_M}{2R}$
d. $-\frac{GM_EM_M}{R}$
e. $-\frac{2GM_EM_M}{R}$

1 answer:

Alex787 [66]2 years ago

7 0

Answer:

option C

Explanation:

given,

mass of earth = Me

mass of the moon = Mm

radius = R

Total mechanical energy = ?

Kinetic energy of moon =

$KE = \dfrac{1}{2}M_m v^2$

velocity of the moon is equal to

$v = \sqrt{\dfrac{GM_e}{R}}$

total energy = $-\dfrac{1}{2}M_m v^2$

now, putting value of in the above equation

$TE = -\dfrac{1}{2}M_m (\sqrt{\dfrac{GM_e}{R}})^2$

$TE = -\dfrac{1}{2}M_m\dfrac{GM_e}{R}$

$TE = -\dfrac{1}{2}\dfrac{GM_mM_e}{R}$

$TE = -\dfrac{GM_mM_e}{2R}$

hence, the correct answer is option C

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Answer:

please read the answer below

Explanation:

The angular momentum is given by

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By taking into account the angles between the vectors r and v in each case we obtain:

a)

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angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

b)

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angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

c)

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angle = 0°

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L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

$L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}$

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

$L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}$

h)

r=(1,-1)

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the same as g):

L = 5kgm/s

hope this helps!!

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An airplane is delivering food to a small island. It flies 100 m above the ground at a speed of 150 m/s .

miss Akunina [59]

Answer:

The airplane should release the parcel $6.7*10^2$ m before reaching the island

Explanation:

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$y=y_0 - \frac{gt^2}{2}$ [1]

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Answer:

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Upon reaching the floor, the first ball (basketball) collides with the floor, this process is very fast, at the end of the process the basketball comes out with a velicad up and collides with the much lighter tennis ball that is still descending .

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Here we must separate the system, creating a system for the conservation of the energy of the basketball ball and another system for the tennis ball only, the conservation of energy should be applied to each system independently

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