Using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.
so then you sub what you have (v and lambada) in the formula.
then multiply the frequency(f) by the given wavelength and then solve for f
<u>Answer:</u>
<em>Equivalence point and end point are terminologies in pH titrations and they are not the same.
</em>
<u>Explanation:</u>
In a <em>titration the substance</em> added slowly to a solution usually through a pippette is called titrante and the solution to which it is added is called titrand. In acid-base titrations acid is added to base or base is added to acid.the strengths of the <em>acid and base titrated</em> determines the nature of the final solution.
At equivalence point the <em>number of moles of the acid</em> will be equal to the number of moles of the base as given in the equation. The nature of the final solution determines the <em>pH at equivalence point. </em>
<em>A pH less than 7 will be the result if the resultant is acidic and if it is basic the pH will be greater than 7. </em>In a strong base-strong acid and weak base-weak acid titration the pH at the equivalence point will be 7 indicating <em>neutral nature of the solution.
</em>
Answer:
(a) 
(b) 
Explanation:
<u>Given:</u>
= The first temperature of air inside the tire = 
= The second temperature of air inside the tire = 
= The third temperature of air inside the tire = 
= The first volume of air inside the tire
= The second volume of air inside the tire = 
= The third volume of air inside the tire = 
= The first pressure of air inside the tire = 
<u>Assume:</u>
= The second pressure of air inside the tire
= The third pressure of air inside the tire- n = number of moles of air
Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.
Using ideal gas equation, we have

Part (a):
Using the above equation for this part of compression in the air, we have

Hence, the pressure in the tire after the compression is
.
Part (b):
Again using the equation for this part for the air, we have

Hence, the pressure in the tire after the car i driven at high speed is
.
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
brainly.com/question/24136952
#SPJ1
The answers include the following:
- The unit of length most suitable for measuring the thickness of a cell phone is a meter.
- The unit of length most suitable for measuring the height of a backyard tree is a meter.
<h3>What is Meter?</h3>
This is defined as the standard unit for measuring the length of a body and is denoted as m.
Height is a vertical type of length which is why meter was chosen as the most appropriate choice.
Read more about Meter here brainly.com/question/1578784
#SPJ1