Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant
Given:
P = 687 days
A = 1.52 AU
Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³
Answer: 1.3439 x 10⁵ (days²/AU³)
Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final height
is the bomb's initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's final velocity
Knowing this, let's begin with the answers:
<h3>b) Time
</h3>
With the conditions given above, equation (1) is now written as:
(4)
Isolating
:
(5)
(6)
(7)
<h3>a) Final velocity
</h3>
Since
, equation (3) is written as:
(8)
(9)
(10) The negative sign only indicates the direction is downwards
<h3>c) Range
</h3>
Substituting (7) in (2):
(11)
(12)
<span>carrying twice the weight and climbing twice as high</span>
Answer:
vf = 3.27[m/s]
Explanation:
In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.
With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.
With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.
Then we equalize the two stress equations and we can clear the acceleration.
a = 3.58 [m/s^2]
As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.
![x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2Aa%2At%5E%7B2%7D%5C%5C1.5%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A%283.58%29%20%2At%5E%7B2%7D%20%5C%5Ct%20%3D%200.91%20%5Bs%5D)
And the speed can be calculated as follows:
![v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Ba%2At%5C%5Cv_%7Bf%7D%3D0%2B%283.58%2A0.915%29%5C%5Cv_%7Bf%7D%3D%203.27%5Bm%2Fs%5D)
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