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3 years ago

A train travels 225 kilometers in 2.5 hours . what's the trains average speed

2 answers:

8 0

Answer=90 kilometers per hour

225 km 2.5hours

225km/2.5hours=90 kilometers per hour

The average speed is 90 kilometers per hour

Maslowich3 years ago

7 0

Let us first concentrate and write down all the information's already given in the question. Based on the information's the answer to the question can be easily deduced.
Distance traveled by the train = 225 km
Time taken to travel the distance of 225 km = 2.5 hours
Then
Average speed = Distance traveled/Time taken
= 225/2.5 km/hour
= 90 km/hour
So the average speed of the train is 90 kilometers per hour. The procedure for doing this kind of problem is described and it is the simplest way of solving such problems. I hope you have understood the process and got the answer you were looking for.

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3 years ago

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If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

sdas [7]

Data:
$f_{2} = 42 Hz$
n (Wave node)
V (Wave belly)
L (Wave length)
The number of bells is equal to the number of the harmonic emitted by the string.

$f_{n} = \frac{nV}{2L}$

Wire 2 → 2º Harmonic → n = 2

$f_{n} = \frac{nV}{2L}$
$f_{2} = \frac{2V}{2L} 
$
$2V = f_{2} *2L$
$V = \frac{ f_{2}*2L }{2}$
$V = \frac{42*2L}{2}$
$V = \frac{84L}{2}$
$V = 42L$

Wire 1 → 1º Harmonic or Fundamental rope → n = 1

$f_{n} = \frac{nV}{2L}$
$f_{1} = \frac{1V}{2L}$
$f_{1} = \frac{V}{2L}$

If, We have:
V = 42L
Soon:
$f_{1} = \frac{V}{2L}$
$f_{1} = \frac{42L}{2L}$
$\boxed{f_{1} = 21 Hz}$

Answer:

The fundamental frequency of the string:
21 Hz

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