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saveliy_v [14]
1 year ago
13

Which algebraic expression shows the average melting points of helium, hydrogen, and neon if h represents the melting point of h

elium, j represents the melting point of hydrogen, and k represents the melting point of neon?
A. h + j + k
B. hjk
c. h+j+k over 3
d. hjk over 3
Mathematics
2 answers:
Ainat [17]1 year ago
8 0

Answer:

c. h+j+k over 3

Step-by-step explanation:

We know that,

\text{Average}=\frac{\text{Sum of observations}}{\text{Total number of observation}}

Here, h, j and k represents the melting point of helium, hydrogen and Neon respectively,

Since, the sum of melting points = h + j + k,

Also, the total number of gases = 3,

Thus,

\text{Average melting points}=\frac{h+j+k}{3}

Or (h+j+k) over 3

⇒ Option c is correct.

LenKa [72]1 year ago
4 0
Average = mean = sum of the values / number of data

Values = h, j, and k

Number of data = 3

Sum of the values = h + j + k

Mean = [h + j + k] / 3

=> Answer: option c. [h+j+k] over 3
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The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat
Greeley [361]

Answer:

1) L \propto T^2

Using the condition given:

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K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

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Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

L \propto T^2

Using the condition given:

2.205 m = K (3)^2

K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

With K a constant. For this case the period of a pendulumn is given by this general expression:

T = 2\pi \sqrt{\frac{L}{g}}

Where L is the length in m and g the gravity g = 9.8 \frac{m}{s^2}.

Part 2

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

If we square both sides of the equation we got:

T^2 = 4 \pi^2 \frac{L}{g}

And solving for L we got:

L = \frac{g T^2}{4 \pi^2}

Replacing we got:

L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m

Part 3

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

Replacing we got:

T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

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