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Alex_Xolod [135]
3 years ago
9

A violin string that is 50.0 cm long has a fundamental frequency of 440 Hz. What is the

Physics
2 answers:
DENIUS [597]3 years ago
8 0

Answer:

30800 m/s

Explanation:

correct answer

amid [387]3 years ago
5 0
For a standing wave on a string, the wavelength is equal to twice the length of the string:
\lambda=2 L
In our problem, L=50.0 cm=0.50 m, therefore the wavelength of the wave is
\lambda = 2 \cdot 0.50 m = 1.00 m

And the speed of the wave is given by the product between the frequency and the wavelength of the wave:
v=\lambda f = (1.00 m)(440 Hz)=440 m/s
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If the relative humidity is 25 nd the saturation mixing ratio is 24g/kg, what is the mixing ratio?
levacccp [35]

The mixing ratio is 6.

To find the answer, we have to know about the mixing ratio.

<h3>What is mixing ratio?</h3>
  • The mixing ratio must be calculated in a complex manner.
  • A saturated vapor pressure (es) for values of air temperature and an actual vapor pressure (e) for values of dewpoint temperature must be determined in order to determine the mixing ratio.
  • The air temperature and/or dewpoint temperature must first be converted to degrees Celsius (°C) before the vapor pressures can be calculated.
  • The equation below can be used to determine the relative humidity (rh), as well as the actual mixing ratio and saturated mixing ratio,

                         Rh=\frac{w}{w_s} *100

where; w is the mixing ratio and w(s) is the saturation mixing ratio.

  • In our question, it is given that,

                                    Rh=25\\w_s=24

  • Thus, the mixing ratio will be,

                            w=\frac{Rh*w_s}{100} =\frac{25*24}{100}=6

Thus, we can conclude that, the mixing ratio is 6.

Learn more about mixing ratio here:

brainly.com/question/8791831

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5 0
2 years ago
Grace rides her bike at a constant speed of 6 miles per hour. How far can she travel in 2 1/2 hours?
aliina [53]

Answer:

15 miles

Explanation:

6 miles per hour

2 1/2 hours

6 x 2 = 12

6 x 1/2 = 3

12 + 3 = 15

7 0
3 years ago
If Neptune’s mass were reduced, what could be done to maintain the same force of gravitational attraction between Neptune and th
Alenkinab [10]
Make Neptune closer to the sun because then it would have a stronger gravitation pull. Because the closer the objects are, they will have a stronger gravitaional force and when the object has more mass, the gravity is also stronger. So, if the mass is reduced, the gravity force would be reduced, but if you bring neptune closer, the gravity force would increase
5 0
2 years ago
Si el número masico del elemento litio es 7 ¿cual es su nucleo?
sp2606 [1]
El núcleo tiene 3 protones (lo que le da al núcleo una carga de +3, identificándolo como el elemento Litio) y 4 neutrones (lo que le da un número total de masa de 7).
4 0
3 years ago
A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
Maurinko [17]

Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity R_{0}=10\ mCi

Time t_{1}=4\ hours

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

R=R_{0}e^{-\lambda t}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

Put the value into the formula

\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})

\lambda=0.0000154\ s^{-1}

\lambda=1.55\times10^{-5}\ s^{-1}

We need to calculate the half life

Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

(b). We need to calculate the value of N₀

Using formula of N_{0}

N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

Put the value into the formula

N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

4 0
3 years ago
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