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nydimaria [60]
2 years ago
13

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

Physics
1 answer:
pshichka [43]2 years ago
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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Given Vout = 17.33 vpp and R1 = 3 kΩ, find the value of RF required to provide Av = 4.33. (Round your answer to 2 decimal places
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Answer:

The magnitude of V_{2} is 4 V and phase of input voltage is zero

Explanation:

Given:

Output voltage V_{out} = 17.33

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For finding feedback resistance we use gain equation

Gain equation for non inverting op-amp is given by,

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For finding input voltage we use,

   A_{v} = \frac{V_{out} }{V_{2} }

    V_{2} = \frac{17.33}{4.33}

    V_{2} = 4 V

The Phase of V_{2} is zero because output voltage phase is 360°

Therefore, the magnitude of V_{2} is 4 V and phase of input voltage is zero

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3 years ago
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Answer:

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4 0
3 years ago
Read 2 more answers
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