All categories

3 years ago

Explain electric circuits. Be sure to include the features, the roles, and the types. Help PLEASEEEE!!

1 answer:

5 0

If there was any way to do that, then your teacher wouldn't
need to keep you coming into class every day and doing
homework every night. She could just give you the 3 or 4
paragraphs and a few pictures that you're asking me for,
and bada-bing ! you'd know it !

The time it takes, and the amount of homework it takes, is
EXACTLY the time you spent hearing about it in class.

(Unless you're some kind of genius savant prodigy, which
you're not and I'm not.)

You might be interested in

Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar

brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

At State 1:

The first step is to find the vapor pressure

$P_{v1} = \rho_1 P_g_1$

$= \phi_1 P_{x \ at \ 125^0C}$

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

$P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1$

$116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)$

$116 \times 10^3 \times v_{v_1} = 183831.7778$

$v_{v_1} = 1.584 \ m^3/kg$

At State 1:

The next step is to determine the mass of water vapor pressure.

$m_{v1} = \dfrac{V}{v_{v1}}$

$= \dfrac{2.5}{1.584}$

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air $m_a$ $P_{a1} V = m_a \dfrac{\overline R}{M_a}T_1$

$(P_1-P_{v1}) V = m_a \dfrac{\overline R}{M_a}T_1$

$(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)$

$710000= m_a \times 114220.642$

$m_a = \dfrac{710000}{114220.642}$

$m_a = 6.216 \ kg$

For the specific volume $v_{v_1} = 1.584 \ m^3/kg$ , we get the identical value of saturation temperature

$T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)$

$T_{sat} =101.92 ^0\ C$

Thus, at $T_{sat} =101.92 ^0\ C$ , condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

At State 1;

The internal energy is calculated as:

$U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )$

At $T_1$ = 125° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 125 ^0C } = 278.93 + ( 286.16 -278.93) (\dfrac{398-390}{400-390} )$

$=278.93 + ( 7.23) (\dfrac{8}{10} )$

$= 284.714 \ kJ/kg\\$

At $T_1$ = 125° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 125^0C} = u_g = 2534.5 \ kJ/kg$

$U_1 = (m_a u_a \ at \ _{ 125 ^0C }) + ( m_{v1} u_v \ at \ _{125^0C} )$

= 6.216 × 284.714 + 1.578 × 2534.5

= 5768.716 kJ

At State 2:

The internal energy is calculated as:

$U_2 = (m_a u_a \ _{ at \ 110^0 C})+ ( m_{v1} u_v \ _{ at \ 110^0 C} )$

At temperature 110° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 110^0C } = 271.69+ ( 278.93-271.69) (\dfrac{383-380}{390-380} )$

$271.69+ (7.24) (0.3)$

$= 273.862 \ kJ/kg\\$

At temperature 110° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 110^0C}= 2517.9 \ kJ/kg$

$U_2 = (m_a u_a \ at \ _{ 110 ^0C }) + ( m_{v1} u_v \ at \ _{110^0C} )$

= 6.216 × 273.862 + 1.578 × 2517.9

= 5675.57 kJ

Finally, the heat transfer during the process is

Q = U₂ - U₁

Q = (5675.57 - 5768.716 ) kJ

Q = -93.146 kJ

with the negative sign, this indicates that heat is lost from the system.

6 0

3 years ago

A 0.0780 kg lemming runs off a

kotegsom [21]

Answer:

5.01 J

Explanation:

Info given:

mass (m) = 0.0780kg

height (h) = 5.36m

velocity (v) = 4.84 m/s

gravity (g) = 9.81m/s^2

1. First, solve for Kinetic energy (KE)

KE = 1/2mv^2

1/2(0.0780kg)(4.84m/s)^2 = 0.91 J

so KE = 0.91 J

2. Next, solve for Potential energy (PE)

PE = mgh

(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J

so PE = 4.10 J

3. Mechanical Energy , E = KE + PE

Plug in values for KE and PE

KE + PE = 0.91J + 4.10 J = 5.01 J

4 0

2 years ago

Monochromatic light with a wavelength of 384 nm passes through a single slit and falls on a screen 86 cm away. If the distance o

valkas [14]

This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
$y= \frac{m\lambda D}{a}$
from the center of the pattern. In the formula, m is the order of the minimum, $\lambda$ the wavelenght, $D$ the distance of the screen from the slit and $a$ the width of the slit.

In our problem, the distance of the first-order band (m=1) is $y=0.22 cm$ . The distance of the screen is D=86 cm while the wavelength is $\lambda = 384 nm=384 \cdot 10^{-7}cm$ . Using these data and re-arranging the formula, we can find a, the width of the slit:
$a= \frac{m \lambda D}{y}= \frac{1 \cdot 384 \cdot 10^{-7}cm \cdot 86 cm}{0.22 cm}=0.015 cm$

3 0

3 years ago

Under the assumption that the beam is a rectangular cantilever beam that is free to vibrate, the theoretical first natural frequ

BartSMP [9]

Answer:

a) Δf = 0.7 n , e) f = (15.1 ± 0.7) 10³ Hz

Explanation:

This is an error about the uncertainty or error in the calculated quantities.

Let's work all the magnitudes is the SI system

The frequency of oscillation is

f = n / 2π L² √( E /ρ)

where n is an integer

Let's calculate the magnitude of the oscillation

f = n / 2π (0.2335)² √ (210 10⁹/7800)

f = n /0.34257 √ (26.923 10⁶)

f = n /0.34257 5.1887 10³

f = 15.1464 10³ n

a) We are asked for the uncertainty of the frequency (Df)

Δf = | df / dL | ΔL + df /dE ΔE + df /dρ Δρ

in this case no error is indicated in Young's modulus and density, so we will consider them exact

ΔE = Δρ = 0

Δf = df /dL ΔL

df = n / 2π √E /ρ | -2 / L³ | ΔL

df = n / 2π 5.1887 10³ | 2 / 0.2335³) 0.005 10⁻³

df = n 0.649

Absolute deviations must be given with a single significant figure

Δf = 0.7 n

b, c) The uncertainty with the width and thickness of the canteliver is associated with the density

In your expression there is no specific dependency so the uncertainty should be zero

The exact equation for the natural nodes is

f = n / 2π L² √ (E e /ρA)

where A is the area of the cantilever and its thickness,

In this case, they must perform the derivatives, calculate and approximate a significant figure

Δf = | df / dL | ΔL + df /de Δe + df /dA ΔA

Δf = 0.7 n + n 2π L² √(E/ρ A) | ½ 1/√e | Δe

+ n / 2π L² √(Ee /ρ) | 3/2 1√A23 |

the area is

A = b h

A = 24.9 3.3 10⁻⁶

A = 82.17 10⁻⁶ m²

DA = dA /db ΔB + dA /dh Δh

dA = h Δb + b Δh

dA = 3.3 10⁻³ 0.005 10⁻³ + 24.9 10⁻³ 0.005 10⁻³

dA = (3.3 + 24.9) 0.005 10⁻⁶

dA = 1.4 10⁻⁷ m²

let's calculate each term

A ’= n / 2π L² √a (E/ρ A) | ½ 1 /√ e | Δe

A ’= n/ 2π L² √ (E /ρ) | ½ 1 / (√e/√ A) |Δe

A ’= 15.1464 10³ n ½ 1 / [√ (24.9 10⁻³)/ √ (81.17 10⁻⁶)] 0.005 10⁻³

A '= 0.0266 n

A ’= 2.66 10⁻² n

A ’’ = n / 2π L² √ (E e /ρ) | 3/2 1 /√A³ |

A ’’ = n / 2π L² √(E /ρ) √ e | 3/2 1 /√ A³ | ΔA

A ’’ = n 15.1464 10³ 3/2 √ (24.9 10⁻³) /√ (82.17 10⁻⁶) 3 1.4 10⁻⁷

A ’’ = n 15.1464 1.5 1.5779 / 744.85 1.4 10⁴

A ’’ = 6,738 10²

we write the equation of uncertainty

Δf = n (0.649 + 2.66 10⁻² + 6.738 10²)

The uncertainty due to thickness is

Δf = 3 10⁻² n

The uncertainty regarding the area, note that this magnitude should be measured with much greater precision, specifically the height since the errors of the width are very small

Δf = 7 10² n

d) Δf = 7 10² n

e) the natural frequency n = 1

f = (15.1 ± 0.7) 10³ Hz

7 0

3 years ago

1. A block is pulled to the right at constant velocity by a 20N force acting at 30o above the horizontal. If the coefficient of

DiKsa [7]

Answer:

44.6 N

Explanation:

Draw a free body diagram of the block. There are four forces on the block:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

and applied force F pulling right 30° above horizontal.

Sum of forces in the y direction:

∑F = ma

N + F sin 30° − mg = 0

N = mg − F sin 30°

Sum of forces in the x direction:

∑F = ma

F cos 30° − Nμ = 0

F cos 30° = Nμ

N = F cos 30° / μ

Substitute:

mg − F sin 30° = F cos 30° / μ

mg = F sin 30° + (F cos 30° / μ)

Plug in values:

mg = 20 N sin 30° + (20 N cos 30° / 0.5)

mg = 44.6 N

8 0

2 years ago

Read 2 more answers