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galben [10]
3 years ago
5

PLZ HELP!!!!!what would happen if a science experiment if it didn't have a well-defined question(it's for an odessey writer and

this question is worth 26 point plz and thz)
Physics
1 answer:
ella [17]3 years ago
7 0
Well, if you were going to do an experiment without a question (or a well-defined one), you would just be doing one to 'see what happens.' Well, one of two things will occur: Something will happen, or nothing will happen. In either of these cases, if you don't have an original question of why you are doing the experiment, you don't know what you're looking for. Nothing may happen, but how will you know if you weren't really looking for anything to happen. Something may happen, and you may not notice it as you weren't sure what you were looking for. Also, something may happen and you may notice it, but you won't know why it did what it did, or have a theory as to how it could have happened because you didn't try to see if anything specific happened in the first place.

If you're test isn't controlled - or doesn't have a question or hypothesis - then it is basically just an accident. Whatever happens happens, but you aren't doing anything to make anything specific happen.

I don't know what an Odyssey Writer is, but I hope this helps!<span />
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Answer:

The two rays, CY and DM are diverging rays and when extended behind the mirror, they appear to intersect each other at point M'. Therefore, the properties of the images formed here are formed behind the mirror, between the pole and principal focus (f), the images are diminished and are virtual and erect.

Explanation:

<h2>Spherical Mirrors</h2>

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3 years ago
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A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.080
deff fn [24]

Answer:

Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s

Explanation:

When rod is released from rest then due to unbalanced torque about the hinge the system will rotate

Now moment of inertia of the system is given as

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I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}

so we have

I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162

I = 0.02835

now by energy conservation we can say work done by gravity must be equal to change in kinetic energy

so we have

\frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}

\frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)

\omega = 4.32 rad/s

Now speed of 0.08 kg mass when it reaches to bottom point is given as

v = \omega \frac{L}{2}

v = 4.32 (0.45)

v = 1.94 m/s

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