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ryzh [129]
3 years ago
7

1. Is the sequence geometric? If so, identify the common ratio.

Physics
1 answer:
frozen [14]3 years ago
6 0
1) <span>yes;2 6*2=12 12*2=24 24*2=48
2)</span><span>Next Term (or nth term) = ar^n-1 
</span>
a = first term, i.e. 5 
<span>r = common ratio i.e. 3 (as 15/5=3 and 45/15=3 </span>
<span>n = .. </span>
<span>as you already have 1st , 2nd and 3rd terms</span>
<span>substituting now </span>
<span>T4= ar^n-1 </span>
<span>= 5*3^4-1 </span>
<span>= 5*3^3 </span>
<span>= 5*27 </span>
<span>T4 = 135 
</span>T5= ar^n-1 
<span>= 5*3^5-1 </span>
<span>= 5*3^4 </span>
<span>= 5*81 </span>
<span>T5 = 405 </span>


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In the simplified version of Kepler's third law, P 2 = a3, the units of the orbital period P and the semimajor axis a of the ell
Orlov [11]

Answer:

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Explanation:

P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.

Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.

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3 years ago
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Answer:

True

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A 49.0 kg wheel, essentially a thin hoop with radius 0.730 m, is rotating at 114 rev/min. It must be brought to a stop in 22.0 s
belka [17]

Explanation:

Mass of the wheel, m = 49 kg

Radius of the hoop, r = 0.73 m

Initial angular speed of the wheel, \omega_i=114\ rev/min = 11.93\ rad/s

Final angular speed of the wheel, \omega_f=0

Time, t = 22 s

(a) If I is the moment of inertia of the hoop. It is equal to,

I=mr^2

I=49\times (0.73)^2

I=26.11\ kg-m^2

We know that the work done is equal to change in kinetic energy.

W=\Delta E

W=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)

W=-\dfrac{1}{2}\times 26.11\times (11.93^2)

W = -1858.05 Joules

(b) Let P is the average power. It is given by :

P=\dfrac{W}{t}

P=\dfrac{1858.05\ J}{22\ s}

P =84.45 watts

Hence, this is the required solution.

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3 years ago
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2 years ago
Read 2 more answers
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