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Degger [83]
3 years ago
5

A charge is distributed uniformly along a long straight wire. The electric field 2 cm from the wire is 36 N/C. The electric fiel

d 4 cm from the wire is:
Engineering
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

Answer:

New electric field = 18 N/C

Explanation:

Given:

Length (E1) = 2 cm

New length (E2) = 4 cm

Electric field =  36 N/C

Find:

New electric field

Computation:

New electric field = 36 [2 / 4]

New electric field = 36 [1/2]

New electric field = 18 N/C

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Two different fuels are being considered for a 2.5 MW (net output) heat engine which can operate between the highest temperature
sveta [45]

Answer:

If the heat engine operates for one hour:

a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.

b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.

In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.

Explanation:

The Carnot efficiency is obtained as:

\epsilon_{car}=1-\frac{T_c}{T_H}

Where T_c is the atmospheric temperature and T_H is the maximum burn temperature.

For the case (B), the efficiency we will use is:

\epsilon_{b}=0.4\epsilon_{car}

The work done by the engine can be calculated as:

W=\epsilon Q=\epsilon H_v\cdot m_{fuel} where Hv is the heat value.

If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

m=\frac{P\cdot t}{\epsilon H_v}

If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.

TC=m\cdot c

8 0
3 years ago
A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe
vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V_f = V_g = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v_f = 0.001057 m³/kg

v_g = 1.0037 m³/kg

u_f = 486.82 kJ/kg

u_g 2524.5 kJ/kg

h_g = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V_f/v_f + V_g/v_g

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v_g

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m_{out = m₁ - m₂

m_{out = 1.89414  - 0.003985

m_{out = 1.890155 kg

so, Initial internal energy will be;

U₁ = m_fu_f + m_gu_g

U₁ = (V_f/v_f)u_f  + (V_g/v_g)u_g

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E_{in -  E_{out = ΔE_{sys

QΔt - m_{outh_{out = m₂u₂ - U₁

QΔt - m_{outh_g = m₂u_g - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0
3 years ago
The one end of a hollow square bar whose side is (10+N/100) in with (1+N/100) in thickness is under a tensile stress 102,500 psi
netineya [11]

Answer:

The one end of a hollow square bar whose side is (10+N/100) in wit

Explanation:

3 0
3 years ago
Dean is buying a home for $170,000. The mortgage company he decided to use to finance the home requires a 10% down payment. What
ANTONII [103]
They answer is 3. $17,00
8 0
3 years ago
Read 2 more answers
Saturated liquid-vapor mixture of water, called wet steam, in a steam line at 1500 kPa is throttled to 50 kPa and 100°C. What is
frozen [14]

Answer:

x = 0.944

Explanation:

Steam at outlet is an superheated steam, since T > T_{sat}. From steam tables, the specific enthalpy is:

h_{out}=2682.4\,\frac{kJ}{kg}

The throttle valve is modelled after the First Law of Thermodynamics:

h_{in} = h_{out}

Hence, specific enthalpy at inlet is:

h_{in}=2682.4\,\frac{kJ}{kg}

The quality in the steam line is:

x = \frac{2682.4\,\frac{kJ}{kg}-844.55\,\frac{kJ}{kg}}{2791.0\,\frac{kJ}{kg} - 844.55\,\frac{kJ}{kg} }

x = 0.944

6 0
3 years ago
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