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Schach [20]
3 years ago
6

an apple with a mass of 0.95 kilograms hangs from a tree branch 3.0 meters above the ground. what is the potential energy of the

apple?
Mathematics
2 answers:
aleksley [76]3 years ago
5 0

Answer:

Formula for Potential energy is given by:

\text{P.E} = mgh

where,

m is the mass

g is the gravity i.e, g = 9.8 m/s^2

h is the height.

As per the statement:

An apple with a mass of 0.95 kilograms hangs from a tree branch 3.0 meters above the ground.

⇒ m = 0.95 kg and h = 3 meter

Using formula we have;

P.E = mgh

Substitute the given values we have;

P.E = (0.95)(9.8)(3) = 27.930 kg m^2/s^2

USe: \text{1 kg m}^2/{s^2} = J

Therefore,  the potential energy of the apple is, 27.930 J

olga2289 [7]3 years ago
4 0
Egrav=mgh
m=mass
g=gravity
h=height  


0.95*10*3.0=28.5

Egrav=28.5J 
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Answer:

\textsf{1.} \quad y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}\:\:\textsf{ where}\:\:x \geq \dfrac{15-\sqrt{769}}{2}\\\\\quad \textsf{or} \quad y=2\sqrt{x+5}

\textsf{2.} \quad y=-|x+1|+5

Step-by-step explanation:

<h3><u>Question 1</u></h3>


<u>Method 1 - modelling as a quadratic with restricted domain</u>


Assuming that the points given on the graph are points that the <u>curve passes through</u>, the curve can be modeled as a quadratic with a limited domain.  Please note that as the x-intercept has not been defined on the graph, I am not including this in this first method.

Standard form of a quadratic equation:

y=ax^2+bx+c

Given points:

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  • (-1, 4)
  • (4, 6)

Substitute the given points into the equation to create 3 equations:

<u>Equation 1  (-4, 2)</u>

\implies a(-4)^2+b(-4)+c=2

\implies 16a-4b+c=2

<u>Equation 2  (-1, 4)</u>

\implies a(-1)^2+b(-1)+c=4

\implies a-b+c=4


<u>Equation 3  (4, 6)</u>

\implies a(4)^2+b(4)+c=6

\implies 16a+4b+c=6

Subtract Equation 1 from Equation 3 to eliminate variables a and c:

\implies (16a+4b+c)-(16a-4b+c)=6-2

\implies 8b=4

\implies b=\dfrac{4}{8}

\implies b=\dfrac{1}{2}

Subtract Equation 2 from Equation 3 to eliminate variable c:

\implies (16a+4b+c)-(a-b+c)=6-4

\implies 15a+5b=2

\implies 15a=2-5b

\implies a=\dfrac{2-5b}{15}

Substitute found value of b into the expression for a and solve for a:

\implies a=\dfrac{2-5(\frac{1}{2})}{15}

\implies a=-\dfrac{1}{30}

Substitute found values of a and b into Equation 2 and solve for c:

\implies a-b+c=4

\implies -\dfrac{1}{30}-\dfrac{1}{2}+c=4

\implies c=\dfrac{68}{15}

Therefore, the equation of the graph is:

y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}

\textsf{with the restricted domain}: \quad x \geq \dfrac{15-\sqrt{769}}{2}

<u>Method 2 - modelling as a square root function</u>

Assuming that the points given on the graph are points that the <u>curve passes through</u>, and the x-intercept should be included, we can model this curve as a <u>square root function</u>.

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y=\sqrt{x}

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The curve is stretched vertically, so:

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To find a, substitute the coordinates of the given points:

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\implies a=2

\implies a\sqrt{-1+5}=4

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\implies 3a=6

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<h3><u>Question 2</u></h3>

<u>Vertex form of an absolute value function</u>

f(x)=a|x-h|+k

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From inspection of the given graph:

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Substituting the given vertex and the found value of a into the function, the equation of the graph is:

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