How Can You Linearise a Cosine Curve for g-forces in Physics?

  • Thread starter Thread starter Maluras
  • Start date Start date
  • Tags Tags
    Cosine Curve
AI Thread Summary
The discussion revolves around linearizing the formula for g-forces, expressed as g-forces=√(391.88-337.12 cosθ)/9.8. The user seeks to convert this into the linear form y=mx+c but is uncertain whether to treat the angle or its cosine as the independent variable. A key point raised is the concern about the square root of negative values, specifically whether the expression (391.88-337.12 cosθ) can ever be negative. It is clarified that since the range of cos(θ) is between -1 and 1, the expression remains positive, allowing for successful linearization. Understanding this range is crucial for proceeding with the linearization process.
Maluras
Messages
1
Reaction score
0

Homework Statement



For my physics EEI, I have developed the formula: g-forces=√(391.88-337.12 cosθ)/9.8

I need to linearise the graph into the form y=mx+c.

I'm not sure where just the angle is the independent or cos of the angle.

Homework Equations



y=k√(x) can be graphed as y vs. √(x) to linearise

The Attempt at a Solution



The square root of a negative is undefined, therefore √(cosθ) is unable to be done for all values
 
Physics news on Phys.org
Maluras said:

Homework Statement



For my physics EEI, I have developed the formula: g-forces=√(391.88-337.12 cosθ)/9.8

I need to linearise the graph into the form y=mx+c.

I'm not sure where just the angle is the independent or cos of the angle.

Homework Equations



y=k√(x) can be graphed as y vs. √(x) to linearise

The Attempt at a Solution



The square root of a negative is undefined, therefore √(cosθ) is unable to be done for all values
Hello Maluras. Welcome to PF !

Is (391.88-337.12 cosθ ) ever negative?

-1 ≤ cos(θ) ≤ 1

Multiply by -337.12 then add 391.88 .

What's your conclusion regarding the range of (391.88-337.12 cosθ ) ?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top