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aleksley [76]
2 years ago
8

Since many wavelengths of light are measured in nanometers, it's useful to know that planck's constant (h) multiplied with the s

peed of
light (c) is equal to 1240 eVnm.
This is useful, as E = hf, or otherwise written as E= .
With this information in mind, solve the following:
A photon has 1.6 eV of energy. What is the wavelength of this photon?
O a
410 nm
Ob
775 nm
Ос
1984 nm
Od
306 nm
Physics
1 answer:
MrRa [10]2 years ago
4 0

Answer:

E = h f      and since f = c / λ     we have E = h c / λ

Thus  λ = h c / E

Also, 1 ev = 1.6E-19 C

λ = 6.63E-34 * 3.0E8 / (1.6 * 1.6E-19)

λ = 6.63 * 3.0 / (1.6 * 1.6) * E-7

λ = 7.77E-7 = 777 nanometers

(B) is correct

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Five-gram samples of copper and aluminum are at room temperature. Both receive equal amounts of energy due to heat flow. The spe
hammer [34]

Answer:

  • <u><em>c. The copper will get hotter than the aluminum.</em></u>

Explanation:

The <em>energy due to heat flow</em>, i.e. the heat energy, is proportional to the product of the mass, the <em>specific heat capacity</em>, and the change in <em>temperature</em>:

          Heat=mass\times \text{specific heat capacity }\times\Delta T

From which you obtain:

     \Delta T=\dfrac{Heat}{mass\times \text{specific heat capacity}}

That equation tells that the change in temperature is inversely related to the product of the mass and the specific heat.

In the problem, the masses of the samples of copper and aluminum are equal (<em>5.00g</em>) and both samples receive <em>equal amounts of energy due to heat flow</em>, thus the only difference is the <em>specific heat capacity</em> of each sample.

From the above stated relationship between the change in temperature, the heat, the mass, and the specific heat capacity, under the assumption of all the other conditions equal (heat energy and mass), the higher the specific heat capacity the lower the change in temperature, and the lower the specific heat capacity the greater the change in temperature.

The heat capacity of copper (0.09 cal/g°C) is lower than the specific heat capacity of aluminum is (0.22 cal/g°C), thus the increase in temeperature of the copper sample will be greater than that of the aluminum sample. This means that <em>the copper will get hotter than the aluminum (option c.)</em>

5 0
3 years ago
A Level is set up midway between two wood hubs that are about 300 ft. apart. The rod reading on hub A is 9.09 ft. and the readin
Zarrin [17]

Answer:

<em>The correct difference is 3.44 ft</em>

Explanation:

The correct difference is given as

D=R_A-R_B

D=9.09-5.65

D=3.44 ft

4 0
3 years ago
Two sections, A and B, are 0.5 km apart along a 0.05 m diameter rough concrete pipe. A is 115 m higher than B, the water tempera
KonstantinChe [14]

Answer:

Q = 178.41 m^3 / s

Explanation:

Given:

  • Length of the pipe L = 0.5 km
  • Diameter D = 0.05 m
  • Pressure head @ A (P_a / γ )= 21.7 m
  • Pressure head @ B (P_b / γ )= 76.1 m
  • Elevation head Z_a = 115 m
  • Elevation head Z_b = 0 m
  • Minor Losses = 0 m
  • Major Losses = f*L*V^2 / 2*D*g = (500/2*0.05*9.81) *f*V^2 = 509.684*f*V^2
  • Velocity at cross section A and B: V_a = V_b m/s
  • Roughness e = 2.5 mm
  • Dynamic viscosity of water u = 8.9*10^-4 Pa-s
  • Density of water p = 997 kg/m^3

Find:

Flow Rate Q = pi*V*D^2/4  m^3/s  ??

Solution:

We will use the Head Balance as derived from Energy Balance:

(P_a / γ ) - (P_b / γ ) + (V_a^2 - V_b^2) / 2*g + (Z_a - Z_b) = Major Losses

21.7 - 76.1 + 0 + 115-0 = 509.684*f*V^2

f*V^2 = 0.18897199

To find correction factor f which is a function of e / D = 0.05, and Reynold's number which is unknown. In such cases we will guess a value of f and perform iterations as follows:

Guess: f_o = 0.072 (Moody's Chart @ e / D = 0.05 and most turbulent function).

V_o = sqrt(0.18897199 / 0.072) = 1.28504 m/s

Re_o = p*V_o*D / u =  997*1.28504*0.05 / 8.9*10^-4 = 71976.6

1st iteration

f_1 = g (Re_o , e/d) = 0.0718702 (Moody's Chart)

V_1 = sqrt(0.18897199 / 0.0718702) = 1.621528 m/s

Re_1 = p*V_1*D / u =  997*1.621528*0.05 / 8.9*10^-4 = 90823.81698

2nd iteration

f_2 = g (Re_1 , e/d) = 0.0718041 (Moody's Chart)

V_2 = sqrt(0.18897199 / 0.0718041) = 1.662273585 m/s

Re_2 = p*V_2*D / u =  997*1.662273585*0.05 / 8.9*10^-4 = 90865.54854

3rd iteration

f_3 = g (Re_2 , e/d) = 0.0718040  (Moody's Chart)

V_3 = sqrt(0.18897199 / 0.0718040) = 1.622274714 m/s

Re_3 = p*V_3*D / u =  997*1.622274714*0.05 / 8.9*10^-4 = 90865.61182

We can observe the convergence of V to 1.6222 m /s. Hence, the required velocity will be used to calculate the Flow rate Q:

Q =  pi*V*D^2/4 = pi*1.6222*0.05^2 / 4

Q = 178.41 m^3 / s

3 0
3 years ago
Which of the following accurately describes the way in which a muscle moves?
Vaselesa [24]
<h3><u>Answer</u>;</h3>

B. When actin filaments are pulled toward the center of the sarcomere, the fiber shortens.

<h3><u>Explanation;</u></h3>
  • <em><u>The events of muscle fiber shortening occurs with in the sacromeres in the fibers. </u></em>
  • <em><u>Contraction of striated muscle fibers takes place as the sacromeres shorten as myosin heads pull on the actin filaments.</u></em>
  • <em><u>Filament movement starts at the region or zone where thin and thick filaments overlap. </u></em>
  • <em><u>Myofibril contains many sacromeres along its length and thuse myofibrils and muscle cells contract as the sacromeres contract.</u></em>
7 0
3 years ago
Plz i need help for the 5 problems. plz show the work!!!
Artemon [7]

Answer:

1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

8 0
3 years ago
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