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RideAnS [48]
3 years ago
6

An object of mass weighing 5.24 kilograms is raised to a height of 1.63 meters. What is the potential energy of the object at th

at height? Note that acceleration due to gravity is 9.8 meters/second2.
Mathematics
2 answers:
Trava [24]3 years ago
6 0
<span>Find force,
9.8 m/s2 x 5.24 kg = </span><span>51.352 N </span>

Find work,
<span><span>51.352 N x 1.63 m=  </span><span>83.70 N*m


</span>
Example:
To solve this given word problem we can first identify the given and the apt formula to use in this phenomenon: Given: Force = 4, 500 N = 4, 500 kg-m/s^2 Acceleration = 5 m/s^2   </span>
<span>Formula: f=ma </span>
<span>Derivation: m = f/a </span>
<span>Solution: </span><span><span>
1.    </span>M = f/a</span> <span><span>
2.    </span>M = 4,500 kg-m/s^2 / 5 m/s^2</span> <span><span>
3.    </span>M = 900 kg  </span> 

<span>Hence, the object’s mass is </span>900 kg.<span>
</span>



zimovet [89]3 years ago
5 0

Answer:

⇒ 73.7 J = 15 kg × 9.8 m/s² × h

⇒ h = (73.7 J) / ( 15 kg × 9.8 m/s²) = 0.5 m

Step-by-step explanation:

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1)


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2)


\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-2}{1-(-4)}\implies \cfrac{10}{1+4}\implies \cfrac{10}{5}\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=2[x-(-4)]\implies y-2=2(x+4) \\\\\\ y-2=2x+8\implies y=2x+10

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Which percent is equivalent to 3 2/5? 34% is!

↑   ↑   ↑  Hope this helps! :D

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