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vovikov84 [41]
3 years ago
10

How do you determine the wattage capacity needed by a power supply?

Physics
1 answer:
Lady bird [3.3K]3 years ago
6 0
We could use the formula for the Power supply in order to find the wattage capacity and it would be:

P = V²/R or P = V * I

Hope this helps!
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A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative r
sergij07 [2.7K]

Answer:

The workdone is  W = 1.712 *10^{-20 } \  J  

Explanation:

From the question we are told that

    The potential difference is  V  =  107 mV =  107 *10^{-3} \  V

Generally the charge on  Na^{+} is  Q_{Na^{+}} = 1.60 *10^{-19 } \  C

 Generally the workdone is mathematically represented as

         W =  Q_{Na^{+}}V

=>     W = 1.60 *10^{-19 } *  107 *10^{-3}    

=>     W = 1.712 *10^{-20 } \  J    

8 0
3 years ago
How much does coast to coast membership cost?
CaHeK987 [17]
The price of coast to coast membership in united states could lie anywhere between $2,000 to $ 5,000
Unless you're a frequent user of this type of event, i think it would be economically more efficient if you pay the resort on one-day price
8 0
3 years ago
A thin flake of mica (n=1.58) is used to cover one slit of a double-slit interference arrangement. The central point on the wiew
Llana [10]

To develop this problem it is necessary to apply the optical concepts related to the phase difference between two or more materials.

By definition we know that the phase between two light waves that are traveling on different materials (in this case also two) is given by the equation

\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))

Where

L = Thickness

n = Index of refraction of each material

\lambda = Wavelength

Our values are given as

\Phi = 7(2\pi)

L=t

n_1 = 1.58

n_2 = 1

\lambda = 550nm

Replacing our values at the previous equation we have

\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))

7(2\pi) = 2\pi(\frac{t}{\lambda}(1.58-1))

t = \frac{7*550}{1.58-1}

t = 6637.931nm \approx 6.64\mu m

Therefore the thickness of the mica is 6.64μm

3 0
3 years ago
Two engineering students, John with a weight of 92 kg and Mary with a weight of 46 kg, are 30 m apart. Suppose each has a 0.04%
Hoochie [10]

Answer:

F = 6.27 x 10 ¹⁹ N

Explanation:

Given

m₁ = 92 kg, m₂ = 46 kg, % = 0.04% N = 6.022 x 10²³ Z = 18, e = 1.6 x 10 ⁻¹⁹ C, M = 0.018 kg/mol

q₁ = % * [m * N * A * e / M ]  

q₁ = 0.0004 * [ ( 92 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₁ = 3.54 x 10⁶ C

q₂ = 0.0004 * [ ( 46 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₂ = 1.773 x 10⁶ C

Now to determine the electrostatic force con use the equation

F = K * q₁ * q₂ / d²

K = 8.99 x 10 ⁹

F = 8.99 x 10 ⁹ * 3.54 x 10⁶ C * 1.773 x 10⁶ C / (30m)²

F = 6.27 x 10 ¹⁹ N

3 0
3 years ago
Một mẫu vật liệu có khối lượng riêng là 2500kg/m3
Leokris [45]

Explanation:

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2 years ago
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