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Artist 52 [7]
3 years ago
14

If electrical energy costs 7¢/kW•h, calculate

Physics
1 answer:
il63 [147K]3 years ago
4 0

Answer:

10 minutes =(1/6)th of an hour

cost of running it for an hour = 0.09*7=0.56 c

in 10 min the cost should be (1/6)*0.56c=0.08c

Explanation:

Your scanner power is 15W so, it will consume 15*60=90Wh of electricity in an hour. that is 0.09KWh.  

So, cost of running it for an hour = 0.09*7=0.56 c

Which means that in 10 min the cost should be (1/6)*0.56c=0.08c

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When light of wavelength 160 nm falls on a gold surface, electrons having a maximum kinetic energy of 2. 66 ev are emitted. Find
enyata [817]

When the light of wavelength is falling on gold surface, the electrons begin to exchange energies.

a)The work function in eV is Φ =5.097 eV.

b) The cut-off wavelength is λ₀ = 243.71 nm

c) The frequency is ν₀  =1.231 × 10¹⁵ Hz

<h3>What is work function?</h3>

The energy needed for a particle to escape and break through the surface.

The kinetic energy of the light emitted is 2.66 eV and wavelength of the light is 160 nm = 160 × 10⁻⁹ m.

a) The work function of the gold for given maximum kinetic energy is

Φ = hc / λ  - K.Emax

Substituting 6.626 × 10⁻³⁴ J.s for h, 3 × 10⁸ m/s for c and 2.66 eV for K.Emax, work function will be

Φ =8.16 × 10⁻¹⁹ J

1 eV = 1.6 × 10⁻¹⁹

The work function in eV is Φ =5.097 eV.

b) The cutoff wavelength is related to work function as

λ₀ = hc / Φ

Substitute the corresponding values into the equation, we get the cut off wavelength

λ₀ = 243.71 nm

c) The frequency corresponding to the cut-off wavelength is

ν₀ = c / λ₀

Substitute the corresponding values into the equation, we get the frequency,

ν₀  =1.231 × 10¹⁵ Hz

Therefore, the values for the following are

a)The work function in eV is Φ =5.097 eV.

b) The cut-off wavelength is λ₀ = 243.71 nm

c) The frequency is ν₀  =1.231 × 10¹⁵ Hz

Learn more about wave function.

brainly.com/question/17484291

#SPJ4

3 0
2 years ago
how does spatial pattern of heights illustrate the relationship between temperature density and the rate of vertical pressure ch
Anika [276]

The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

Generally, the relationship between temperature, density and rate of vertical pressure is given as;

\rho = \frac{PM}{RT}

\frac{dP}{dz} = -\rho g\\\\

where;

  • <em>ρ is density</em>
  • <em>T is temperature</em>
  • <em>dP is rate of change of vertical  pressure</em>

Thus, from the formula above, we can conclude the following relationship between temperature, density and the rate of vertical pressure change in spatial pattern of heights.

The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

Learn  more here:brainly.com/question/25395377

5 0
3 years ago
You are directed to set up an experiment in which you drop, from shoulder height, objects with similar surface areas but differe
tangare [24]

Answer: c. they will hit the ground at the same time

Explanation:

The volume of both objects is almost the same, so the force of friction will be the same in each one, so we can discard it.

Now, when yo drop an object, the acceleration of the object is always g = 9.8m/s^2 downwards, independent of the mass of the object.

So if you drop two objects with the same volume but different mass, because the acceleration is the same for both of them, they will hit the ground at the same time, this means that the density of the object has no impact in how much time the object needs to reach the floor.

So the correct option is c

3 0
3 years ago
Which sphere of Earth do humans exist in?
PolarNik [594]
Biosphere is the answer, hope i helped :))
4 0
2 years ago
Read 2 more answers
What minimum distance should you separate two sources emitting the same waves with wavelength 5mm in phase such that you obtain
Makovka662 [10]

To solve this problem we will apply the concept related to destructive interference (from the principle of superposition). This concept is understood as a superposition of two or more waves of identical or similar frequency that, when interfering, create a new wave pattern of less intensity (amplitude) at a point called a node. Mathematically it can be described as

d = n \frac{\lambda}{2}

Where,

d = Path difference

\lambda= wavelength

n = Any integer which represent the number of repetition of the spectrum

In this question the distance between the two source will be minimum for the case of minimum path difference, then n= 1

d = \frac{\lambda}{2}

d = \frac{5*10^{-3}}{2}

d = 2.5mm

Therefore the minimum distance that should you separate two sources emitting the same waves is 2.5mm

8 0
3 years ago
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