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omeli [17]
4 years ago
7

% = Wo/Wi x 100 Solve for Wo % = Wo/Wi x 100 Solve for Wi

Physics
1 answer:
irga5000 [103]4 years ago
8 0
1) % = (Wo /Wi) * 100

Solve for Wo => Wo = (% / 100) * Wi

For example, % =30% and Wi = 250 => Wo = (30 /100) * 250 = 0.30 * 250 = 75

Wo = 75

2) % = (Wo / Wi) * 100

Solve for Wi

=> Wi = Wo * (%/100)

For example, Wo = 125 and % = 40%

=> Wi = 125 * (40 / 100) = 125 * 0.40 = 50

Wi = 50
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Anna [14]
Pitch is the sensation of certain frequencies to the ear. High frequency = high pitch, low frequency = low pitch. 

f = c(speed of the wave) /  <span>λ (wavelength)

1. 343m/s / 0.77955m = 439.99 Hz   
     This corresponds to pitch A 

2. 343m/s / 0.52028m = 659.26 Hz
</span>     This corresponds to pitch E 
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3. 343m/s / 0.65552m = 523.349 Hz
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  λ = c / f<span>
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3 0
3 years ago
Which of these processes in the water cycle occur at a very high rate, to cause a hurricane?
Pavel [41]
Answer: d. evaporation and condensation

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5 0
4 years ago
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

5 0
3 years ago
A constant magnetic field can be used to produce an electric current. True or False?
Umnica [9.8K]

this question is true

6 0
4 years ago
A kitchen mixer requires 375 W of electric power. If a voltage of 125 V is available to operate the mixer, what is the required
jenyasd209 [6]
Watts=V*I so in turn I= W/V 375/125 = 3
It would take 3 Amps
6 0
3 years ago
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