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4 years ago

A standard for comparison that helps to ensure that the experimental result is caused by the condition being tested is the ____.

1 answer:

4 0

A standard for comparison that helps to ensure that the experimental result is caused by the condition being tested is the control.

Explanation:

In an experiment, we need to use a reference or standard term with which we compare our obtained results for better accuracy and precision. So there will several conditions or boundary conditions that need to include in those experiments to prevent the experiment from having a fallback or repair.

Among these several conditions, some of the conditions are mostly given to compare with the conditions occurred during the running of the experiment. And if any case, there is any change in those conditions with respect to standard conditions during the running of the experiment, then the experiment should be stopped.

So, these conditions which are standard comparison and helps to ensure that the experimental result is caused by the condition being tested and not anything else is termed as control.

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3 years ago

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Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

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b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

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