the calculated frequency is 932 Hz.
Even for sensors positioned at significant track offsets, the computation method integrates three different speed calculation techniques to produce estimates for arbitrary train speeds. The second technique is comparable, but it combines a running rms with a previously created "dominant frequency method." The third technique calculates train speed using regression analysis and an analytical vibration frequency prediction model.
given-
train is travelling at a constant speed. f=(π v/L)
The actual frequency of the note emitted by the train.
By applying Doppler effect,
n'=n( v + vs/ v-vs )
=200( 340+220 / 340-220 )
=200( 560/120)
=932 Hz
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2H2O2 ------> 2H2O +O2
because we're given one side with 2 hydrogens and 2 oxygens and another sie with 3 oxygens and 2 hydrogens. The hydrogens are balanced but the oxygens are not. You can't make the oxygen a three on the reactants side with a whole number so you multiply by 2.
Now you have 4 hydrogens and 4 oxygens on one side and 4 hydrogens and 4 oxygens on the other :)
<span>C. market development
</span>The process of naming broad product-markets and then dividing them in order to select target markets and develop suitable marketing mixes is called:<em><u>
MARKET DEVELOPMENT
NOT:
</u></em><em><u /></em><u />A. market penetration
B. market segmentation
<span>D. market research</span><em><u>
</u></em>
Answer:
6 bricks
Explanation:
For the bricks system to NOT fall over, then the center of mass of the system must lie within the touching area of the bottom brick.
Let the reference line be at the left end of the base brick, and the bricks are being stacking to the right direction.
- The 1st brick would have a center of mass at 20/2 = 10 cm or 10 + 0, this is < 20 cm so it stays
- 2 stacked bricks would have a center of mass at (10 + 14)/2 = 12 cm or 10 + 2, this is also < 20 cm so it stays
- 3 stacked bricks would have a center of mass at (10 + 14 + 18) / 3 = 14 cm or 10 + 2*2, this is also < 20 cm so it stays
- 4 stacked bricks would have a center of mass at (10 + 14 + 18 + 22)/4 = 16 cm or 10 + 2*3, this is also < 20 cm so it stays
- 5 stacked bricks would have a center of mass at (10 + 14 + 18 + 22 + 26)/5 = 18 cm or 10 + 2*4, this is also < 20 cm so it stays
- 6 stacked bricks would have a center of mass at (10 + 14 + 18 + 22 + 26 + 28)/5 = 20 cm or 10 + 2*5, this is also <= 20 cm so it stays before tipping over.
- The 7th brick would make everything fall down.
B should be the answer
Look at this example to help you
