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e-lub [12.9K]
3 years ago
12

A rock falls from rest a vertical distance of 0.72 meter to the surface of a planet in 0.63 second. What is the magnitude of the

acceleration due to gravity on the planet?
Mathematics
1 answer:
Reika [66]3 years ago
6 0
The question is asking to calculate the magnitude of the acceleration due to gravity on the planet and base on the given of the problem and in my further calculation, the magnitude would be 3.638 m/s^2. I hope you are satisfied with my answer and feel free to ask for more 
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the pitch, or frequency, of a vibrating string varies directly with the square root of the tension. if a string vibrates at a fr
Naily [24]
\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad  y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------\\\\
\begin{array}{llll}
\stackrel{frequency}{f}\textit{ of a vibrating string varies directly}\\
\qquad \qquad \textit{with the square root of the tension }\stackrel{tension}{t}
\end{array}

\bf f=k\sqrt{t}\quad \textit{we also know that }
\begin{cases}
f=300\\
t=8
\end{cases}\implies 300=k\sqrt{8}
\\\\\\
\cfrac{300}{\sqrt{8}}=k\implies \cfrac{300}{\sqrt{2^2\cdot 2}}=k\implies \cfrac{300}{2\sqrt{2}}=k\implies \cfrac{150}{\sqrt{2}}=k
\\\\\\
\textit{and we can \underline{rationalize} it to }\cfrac{150\sqrt{2}}{2}\implies 75\sqrt{2}=k

\bf thus\qquad f=\stackrel{k}{75\sqrt{2}}\sqrt{t}\implies \boxed{f=75\sqrt{2t}}\\\\
-------------------------------\\\\
\textit{now, when t = 72, what is \underline{f}?}\qquad f=75\sqrt{2(72)}
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Can someone Help with this question
inn [45]

Answer:

B 12

Step-by-step explanation:

  • manuel ate 1/3X
  • His brother ate 1/4X
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manuel+brother+leftover=X

x/3+x/4+5=x

7x+60/12=x

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7x+60=12x

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60=5x

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x=60/5

x= 12

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Real-all numbers
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