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2 years ago

How does energy transition from one form to another as water moves from behind a dam to downstream of a dam?.

Engineering

1 answer:

Tresset [83]2 years ago

3 0

Answer:
potential energy changes to kinetic energy
Explanation:

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What were the trade-offs and opportunity costs of one of Teasha’s economic decisions?

sergij07 [2.7K]

Answer:

Opportunity cost she loses the chance to get the degree. The trade- off is she gets to start a business.

Explanation:

3 0

3 years ago

The equilibrium fraction of lattice sites that are vacant in electrum (a silver gold alloy) at 300K C is 9.93 x 10-8. There are

IrinaK [193]

Answer:

the density of the electrum is 14.30 g/cm³

Explanation:

Given that:

The equilibrium fraction of lattice sites that are vacant in electrum = $9.93*10^{-8}$

Number of vacant atoms = $5.854 * 10^{15} \ vacancies/cm^3$

the atomic mass of the electrum = 146.08 g/mol

Avogadro's number = $6.022*10^{23$

The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)

$5.854*10^{15}$ = $9.93*10^{-8}$ × Total number of sites(N)

Total number of sites (N) = $\dfrac{5.854*10^{15}}{9.93*10^{-8}}$

Total number of sites (N) = $5.895*10^{22}$

From the expression of the total number of sites; we can determine the density of the electrum;

$N = \dfrac{N_A \rho _{electrum}}{A_{electrum}}$

where ;

$N_A$ = Avogadro's Number

$\rho_{electrum} =$ density of the electrum

$A_{electrum}$ = Atomic mass

$5.895*10^{22} = \dfrac{6.022*10^{23}* \rho _{electrum}}{146.08}$

$5.895*10^{22} *146.08}= 6.022*10^{23}* \rho _{electrum}$

$8.611416*10^{24}= 6.022*10^{23}* \rho _{electrum}$

$\rho _{electrum}=\dfrac{8.611416*10^{24}}{6.022*10^{23}}}$

$\mathbf{ \rho _{electrum}=14.30 \ g/cm^3}$

Thus; the density of the electrum is 14.30 g/cm³

7 0

3 years ago

A 75 ohm coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 + j75 Ohm. If the diel

Hatshy [7]

Answer:

The load reflection coefficient, $\Gamma =0.62\angle 82.875^{\circ} \Omega$

Reflection coefficient at input, $\Gamma = 0.62\angle - 147.518^{\circ} \Omega$

SWR = 4.26

Given:

Characteristic impedance of the co-axial cable, $Z_{c} = 75 \Omega$

Length of the cable, L = 2.0 cm = 0.02 m

$Z_{Load} = 37.5 + j75 \Omega$

Dielectric constant, K = 2.56

frequency, f = 3.0 GHz = $3.0 \times 10^{9} Hz$

Explanation:

In order to calculate the reflection coefficient at load, we first calculate these:

The line input impedance $Z_{i}$ is given by:

$Z_{i} = Z_{c}\frac{Z_{Load} + jZ_{c} tan(\beta L)}{Z_{c} + jZ_{Load} tan (\beat L)}$ (1)

Now, we calculate the value of $\beta$ :

$\beta = \frac{2\pi}{\lambda'} = \farc{2\pi f\sqrt{K}}{c}$

(since, $\lambda' = \farc{c}{f\sqrt{K}}$ )

$\beta = \farc{2\pi f\sqrt{2.56}}{3\times 10^{8}} = 100.53$

Now, Substituting the value in eqn (1):

$Z_{i} = 75\frac{37.5 + j75 + j75 tan(100.53\times 0.02)}{75 + j(37.5 + j75) tan ( 100.53\times 0.02)} = 18.99 - j20.55 \Omega = 27.98\angle - 47.257^{\circ} \Omega$

Now, the load reflection coefficient is given by:

$\Gamma = \frac{Z_{Load} - Z_{c}}{Z_{c} + Z_{Load}}}$

Thus

$\Gamma = \frac{37.5 + j75 - 75}{75 + 37.5 + j75}} = 0.077 + j0.615 = 0.62\angle 82.875^{\circ} \Omega$

Similarly,

Reflection coefficient at input:

$\Gamma' = \frac{Z_{i} - Z_{c}}{Z_{c} + Z_{i}}}$

$\Gamma' = \frac{18.99 - j20.55 - 75}{75 + 18.99 - j20.55}} = - 0.523 - j0.334 = 0.62\angle - 147.518^{\circ} \Omega$

Now, the SWR is given by:

SWR, Standing Wave Ratio = $\frac{1 +|\Gamma|}{1 - |\Gamma|}$

SWR = $\frac{1 +|0.62|}{1 - |0.62|} = 4.26$

8 0

3 years ago

A charge of 11.748 nC is uniformly distributed along the x-axis from −2 m to 2 m . What is the electric potential (relative to z

svp [43]

Answer:

electric potential = 22.36 volt

Explanation:

given data

charge Q = 11.748 nC

distance d = 5 - 2 = 3 m

length = 2 + 2 = 4 m

Coulomb constant = 8.98755 × 109 N·m²/C ²

solution

electric potential is express as

electric potential = $\frac{Q}{4\pi \epsilon _o L}$ $ln(1+\frac{L}{d})$ ..............1

electric potential = $\frac{KQ}{L} ln(1+\frac{L}{d})$

put here value

electric potential = $\frac{8.98755\times 10^9\times 11.748\times 10{-9}}{4} ln(1+\frac{4}{3})$

electric potential = 22.36 volt

6 0

3 years ago

Which of the following circumstances call for a greater than normal following distance?

mariarad [96]

Explanation:

The three-second rule is recommended for passenger vehicles during ideal road and weather conditions. Slow down and increase your following distance even more during adverse weather conditions or when visibility is reduced. Also increase your following distance if you are driving a larger vehicle or towing a trailer.

7 0

3 years ago