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irinina [24]
3 years ago
10

Asteroid Ida was photographed by the Galileo spacecraft in 1993, and the photograph revealed that the asteroid has a small moon,

which has been named Dactyl. Assume the mass of Ida is 4.4 x 1016 kg, the mass of Dactyl is 2.6 x 1012 kg, and the distance between the center of Dactyl and Ida is 95 km. G = 6.672x10-11 N-m2/kg2 Part Description Answer Save Status A. Assuming a circular orbit, what would be the orbital speed of Dactyl? (include units with
Physics
1 answer:
Nady [450]3 years ago
3 0

Answer:

The orbital speed of Dactyl is 5.55m/s

Explanation:

The orbital speed can be determined by the combination of the universal law of gravity and Newton's second law:

F = G\frac{M \cdot m}{r^{2}}  (1)

Where G is gravitational constant, M is the mass of the asteroid, m is the mass of the moon and r is the distance between them

In the other hand, Newton's second law can be defined as:

F = ma  (2)

Where m is the mass and a is the acceleration

Then, equation 2 can be replaced in equation 1

m\cdot a  = G\frac{M \cdot m}{r^{2}}  (2)

However, a will be the centripetal acceleration since the moon Dactyl describe a circular motion around the asteroid

a = \frac{v^{2}}{r}  (3)

m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}} (4)

Therefore, v can be isolated from equation 4:

m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r

m \cdot v^{2} = G \frac{M \cdot m}{r}

v^{2} = G \frac{M \cdot m}{rm}

v^{2} = G \frac{M}{r}

v = \sqrt{\frac{G M}{r}} (5)

Finally, the orbital speed can be found from equation 5:

Notice, that it is necessary to express r in units of meters.

r = 95km \cdot \frac{1000m}{1km} ⇒ 95000m

v = \sqrt{\frac{(6.672x10^{-11}N.m^{2}/kg^{2})(4.4x10^{16}kg)}{95000m}}

v = 5.55m/s

Hence, the orbital speed of Dactyl is 5.55m/s

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If a quasar is moving away from us at v/c = 0.8, what is the measured redshift?
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Answer:

The measured redshift is z =2

Explanation:

Since the object is traveling near light speed, since v/c = 0.8, then we have to use a redshift formula for relativistic speeds.

z= \sqrt{\cfrac{c+v}{c-v}}-1

Finding the redshift.

We can prepare the formula by dividing by lightspeed inside the square root to both numerator and denominator to get

z= \sqrt{\cfrac{1+\cfrac vc}{1-\cfrac vc}}-1

Replacing the given information

z= \sqrt{\cfrac{1+0.8}{1-0.8}}-1

z= \sqrt{\cfrac{1.8}{0.2}}-1\\z= \sqrt{9}}-1\\z=3-1\\z=2

Thus the measured redshift is z = 2.

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3 years ago
How much time will it take to perform 440 joule of work at a rare of 11 w?​
kifflom [539]

Answer:

40sec

Explanation:

Data

Work = 440 J

Power= 11watt

time = ?

Power = work done/time

===> time = work done/power

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2 years ago
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PilotLPTM [1.2K]

Answer:

a) A = 3 cm,  b)  T = 0.4 s,   f = 2.5 Hz,

2) A standing wave the displacement of the wave is canceled and only one oscillation remains

Explanation:

a) in an oscillatory movement the amplitude is the highest value of the signal in this case

          A = 3 cm

b) the period of oscillation is the time it takes for the wave to repeat itself in this case

          T = 0.4 s

the period is the inverse of the frequency

         f = 1 /T

         f = 1 /, 0.4

         f = 2.5 Hz

2) a traveling wave is a wave for which as time increases the displacement increases, in the case of a transverse wave the oscillation is perpendicular to the displacement and in the case of a longitudinal wave the oscillation is in the same direction of the displacement.

A standing wave occurs when a traveling wave bounces off some object and there are two waves, one that travels in one direction and the other that travels in the opposite direction. In this case, the displacement of the wave is canceled and only one oscillation remains.

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I think A

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Read 2 more answers
A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Determine the
marusya05 [52]

Answer:

Explanation:

Given

Initial velocity of ball u=10\ m/s

height of window h=20\ m

Using Equation of motion

y=ut+\frac{1}{2}at^2

where u=initial velocity

t=time

a=acceleration

As ball is already is at a height of 20 m so

Y=ut+\frac{1}{2}at^2+20

Y=10\times t+0.5\times (-9.8)t^2+20

Y=-4.9t^2+10t+20

(b)highest point is obtained at v=0

v^2-u^2=2as

where

v=final velocity

u=initial velocity

a=acceleration

s=displacement

(0)-10^2=2\times (-9.8)\times s

s=\frac{100}{19.6}

s=5.102\ m

Highest Point will be s+20=25.102\ m

(c)Time taken when the ball hit the ground i.e. at Y=0

-4.9t^2+10t+20=0

t=3.28\ s

impact velocity v=\sqrt{2\times 9.8\times 25.102}

v=22.181\ m/s

7 0
3 years ago
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