The straight line in the experiment of hooks law does not pass through the origin beacause?

1 answer:

7 0

Okay, let’s look at it this way: when does a line pass through the origin? The line represents possible values of x and y that satisfy the equation.

So, when a line passes through the origin, it passes through the coordinates (0,0). x = 0, y = 0. So, let’s model this with the equation ax + by + c = 0. Sub in x = 0 and y = 0 to the equation and we find that 0 + 0 + c = 0. Clearly, c = 0.

So, with this simple explanation, I hope you understand when a line does pass through the origin.

Now, let’s look at when a line doesn’t pass through the origin. This is when c is not equal to 0. Hence, when x = 0, y cannot equal 0; c + by = 0, and we know that c is not 0. If y is 0, then we get c = 0… where c is not 0. Ehh. Thus, you can see that a line does not pass through the origin when c is not equal to 0 by ehat is hope is a simple explanation. You don’t need to know how to prove it, I presume, but that’s not too hard either.

Oops, I realised that I just assumed you were talking about linear graphs. For quadratic graphs, the reasoning is similar. For graphs of the form y = ax^2, the minimum/maximum point of the graph will be the origin. For graphs of the form y = ax^2 + bx, it will pass through the origin but the line of symmetry will be different. For graphs of the form y = ax^2 + bx + c (you know, where c is not zero) , the graph will not pass through the origin because the maximum/ minimum point is actually raised or lowered by c units

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A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of

kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

The mass of the pole vaulter, m = 57 kg

The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

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