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2 years ago

How many moles are in 73.4 grams of Phosphorus?

Physics

1 answer:

mylen [45]2 years ago

4 0

23.2

hope this helped

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What is the target heart rate for a 24 year old?

andriy [413]

Answer:

196

Explanation:

subtract 24 from 220 to get your answer.

6 0

2 years ago

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One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

3 years ago

Most paper does not reflect light very well because its surface is somewhat rough.<br><br> T or F

Sloan [31]

I think the correct answer is true. Most paper does not reflect light very well because its surface is somewhat rough. Light only reflects to surfaces which has a smooth texture or have a uniform texture on the surface. Hope this answers the question. Have a nice day.

3 0

3 years ago

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What was Marie's speed from point A to point B, in miles/minute?

sveticcg [70]

Answer:

40m

Explanation:

40m

6 0

2 years ago

Inserting the formulas you found for Xman(t) and Xbus(t) into the conditionXman(tcatch)=Xbus(tcatch) , you obtain the following-

lara [203]

Answer:

c > √(2ab)

Explanation:

In this exercise we are asked to find the condition for c in such a way that the results have been real

The given equation is

½ a t² - c t + b = 0

we can see that this is a quadratic equation whose solution is

t = [c ±√(c² - 4 (½ a) b)] / 2

for the results to be real, the square root must be real, so the radicand must be greater than zero

c² -2a b > 0

c > √(2ab)

3 0

2 years ago