Ask question

Ask question

All categories

3 years ago

13

A mysterious device found in a forgotten laboratory accumulates charge at a rate specified by the expression gm = 9 - 10tC from

the moment it is switched on. (a) Calculate the total charge contained in the device at t: O. (b) Calculate the totalcharge contained at t: 1 s. (c) Determine the current flowing into the device at t: 1 s, 3 s, and 10 s.

1 answer:

Katyanochek1 [597]3 years ago

5 0

Answer:

9C

-1C

I (1) = I (3) = I(10) = -10 Amps

Explanation:

Given expression : q(t) = 9 - 10*t

a) @ t = 0, q(0) ?

q (0) = 9 - 10 * 0 = 9 C

Answer : 9C

b) @ t = 1, q(1) ?

q (1) = 9 - 10 * 1 = 1 C

Answer : 1C

c) t = 1 , 3 , 10 s what is I (1) , I (3), I (10) ?

I = dq / dt

I = dq/dt = -10 Amps

Hence,

I (1) = I (3) = I(10) = -10 Amps

Answer : I (1) = I (3) = I(10) = -10 Amps

You might be interested in

Which of the following is not a function of the suspension system?_____

tatyana61 [14]

The awnser is D cause it is

4 0

3 years ago

A motor cycle is moving up an incline of 1 in 30 at a speed of 80 km/h,and then suddenly the engine shuts down.The tractive resi

Ad libitum [116K]

Explanation:

option iii is the right answer.

4 0

3 years ago

P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame

eimsori [14]

Answer:

a) $m=0.17kg/s$

b) $Ma=0.89$

Explanation:

From the question we are told that:

Pressure $P=60kPa$

Diameter $d=3cm$

Generally at sea level

$T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4$

Generally the Power series equation for Mach number is mathematically given by

$\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}$

$\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}$

$Ma=0.89$

Therefore

Mass flow rate

$\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\rho=0.848kg/m^3$

Generally the equation for Velocity at throat is mathematically given by

$V=Ma(r*T_0\sqrt{T_e}$ )

Where

$T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}$

$T_e=248$

Therefore

$V=0.89(1.4*288\sqrt{248})\\\\V=284$

Generally the equation for Mass flow rate is mathematically given by

$m=\rho*A*V$

$m=0.84*\frac{\pi}{4}*3*10^{-2}*284$

$m=0.17kg/s$

6 0

3 years ago

A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is mainta

strojnjashka [21]

Answer:

$\dot W_{out} = 399.47\,kW$

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0$

The work done by the turbine is:

$\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}$

The properties of the water are obtained from property tables:

Inlet (Superheated Steam)

$P = 10\,MPa$

$T = 520\,^{\textdegree}C$

$h = 3425.9\,\frac{kJ}{kg}$

Outlet (Superheated Steam)

$P = 1\,MPa$

$T = 280\,^{\textdegree}C$

$h = 3008.2\,\frac{kJ}{kg}$

The work output is:

$\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW$

$\dot W_{out} = 399.47\,kW$

5 0

3 years ago

Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

$h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k$

Obtain the following properties at 10kPa from the table "saturated water"

$h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K$

Calculate the enthalpy at exit of the turbine using the energy balance equation.

$\frac{dE}{dt}=Q-W+m(h_1-h_2)$

Since, the process is isentropic process $Q=0$

$0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg$

Use the isentropic relations:

$s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87$

Calculate the enthalpy at isentropic state 2s.

$h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg$

a.)

Calculate the isentropic turbine efficiency.

$\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%$

b.)

Find the quality of the water at state 2

since $h_f$ at 10KPa < $h_2$ < $h_g$ at 10KPa

Therefore, state 2 is in two-phase region.

$h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9$

Calculate the entropy at state 2.

$s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K$

Calculate the rate of entropy production.

$S=\frac{Q}{T}+m(s_2-s_1)$

since, Q = 0

$S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k$

6 0

4 years ago