How would you express the frequency of a wave in which five peaks passed a given point each second?

Chemistry

1 answer:

Pani-rosa [81]3 years ago

3 0

n = number of peaks passed a particular point = 5

t = time taken by the peaks to pass a particular point = 1 sec

f = frequency of the wave

by definition , frequency is the repetition of an event in a given interval of time.

we know that frequency is given as

f = n/t

inserting the values

f = 5/1

f = 5 Hz

hence the frequency of the wave comes out to be 5 Hz

You might be interested in

Can someone help me sort this :(

jeyben [28]

Answer: A. is prey B. is predator C. is habitat D. is comunity E. is plants are producers and plant eating animals consumors F. is food chain G. is food web H. parasite I decomposers and scavengers J. is compete K an adsption

Explanation:

7 0

3 years ago

A sample of propane (C3H8) has a mass of 0. 47 g. The sample is burned in a bomb calorimeter that has a mass of 1. 350 kg and a

Nady [450]

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The specific heat has been expressed as:

$q=mc\Delta T$

<h3 /><h3>Computation for the heat absorbed</h3>

The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.

$q_{released}=q_{absorbed}\\
q_{released}=m_{calorimeter}\;c_{calorimeter}\;\Delta T$

The given mass of calorimeter has been, $m_{calorimeter}=1350\;\rm g$

The specific heat of the calorimeter has been, $c_{calorimeter}=5.82\;\rm J/g^\circ C$

The change in temperature of the calorimeter has been, $\Delta T=2.87^\circ \rm C$

Substituting the values for heat released:

$q_{released}= 1350\;\text g\;\times\;5.82\;\text J/\text g^\circ \text C\;\times\;2.87^\circ \text C\\
q_{released}=22,549.5\;\text J\\
q_{released}}=22.54\;\rm kJ$