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3 years ago

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After the NEAR spacecraft passed Mathilde, on several occasions rocket propellant was expelled to adjust the spacecraft's moment

um in order to follow a path that would approach the asteroid Eros, the final destination for the mission. After getting close to Eros, further small adjustments made the momentum just right to give a circular orbit of radius 45 km (45 × 103 m) around the asteroid. So much propellant had been used that the final mass of the spacecraft while in circular orbit around Eros was only 545 kg. The spacecraft took 1.04 days to make one complete circular orbit around Eros. Calculate what the mass of Eros must be:

1 answer:

olya-2409 [2.1K]3 years ago

5 0

Mamaste cuzzo alv el mundo #TrokitasDelValle956

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How do I go about this?

Anna71 [15]

Hi there!

(a)

Recall that:
$W = F \cdot d = Fdcos\theta$

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

$W =248(56)cos(30) = 12027.36 J$

To the nearest multiple of ten:
$W_A = \boxed{12030 J}$

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
$\boxed{W_g = 0 J}$

(c)
Similarly, the normal force is perpendicular to the displacement, so:
$\boxed{W_N = 0 J}$

(d)

Recall that the force of kinetic friction is given by:
$F_{f} =\mu_k mg$

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
$W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J$

In multiples of ten:
$\boxed{W_f = -3070 J}$

(e)
Simply add up the above values of work to find the net work.

$W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}}$

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
$F_{net} = F_{Ax} - F_f$

$W = F_{net} \cdot d = (F_{Ax} - F_f)$

$W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}$

5 0

2 years ago

11. You want to calculate the displacement of an object thrown over a bridge. Using -10 m/s2 for acceleration due to gravity, wh

Natalka [10]

Displacement is d

Vf² = Vi² + 2 g d

(-20²) = (+10²) + 2 (-9.8) d

-19.6 d = 300

d = -15.3 m

negative means lower

time is t

d = Vi t + 1/2 g t²

-15.3 = 10 t + (-4.9) t²

4.9 t² - 10 t -15.3 = 0

t = 3.06 s

3 0

3 years ago

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each othe

deff fn [24]

Answer:

$\theta_2 - \theta_1 = 156.93 degree$

Explanation:

As we know that the displacement of the particle from the mean position is 1/5 times of its amplitude

so we have

$y = A sin\omega t$

$y = \frac{A}{5}$

so now we have

$\frac{A}{5} = A sin\omega t$

now we have

$\theta_1 = 11.53 degree$

so the phase other particle in opposite direction is given as

$\theta_2 = 180 - 11.53 = 168.46 degree$

so we have phase difference given as

$\theta_2 - \theta_1 = 168.46 - 11.53$

$\theta_2 - \theta_1 = 156.93 degree$

7 0

3 years ago

Please Help! 30 points! ASAP PLZ :)

Savatey [412]

The best logical answer is A

5 0

3 years ago

Read 2 more answers

Hen a gfci receptacle device is installed on a 20-ampere branch circuit (12 awg copper), what is the minimum volume allowance (i

Bezzdna [24]

Answer:

2.25in³

Explanation:

For a 12 awg conductor the minimum volume allowance as stated by the NEC is 2.25in³

See attached Table 314.16(B) from NEC 2011

4 0

3 years ago