How long will it take a Honda Civic to travel 118 miles if it is travelling at an average speed of 72 mph?

Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor. (a) Find the charge and energy stored if the ca

Andru [333]

Answer:

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

Explanation:

a)

Identify the unknown: 

The charge and energy stored if the capacitors are connected in series

List the Knowns:

Capacitance of the first capacitor: $C_{1}$ = 2цF = 2 x 10-6 F

Capacitance of the second capacitor $C_{2}$ = 7.4цF = 7.4 x 10-6 F

Voltage of battery: V = 9 V

Set Up the Problem:

Capacitance of a series combination:

$\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }$ +............

$\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F\\$

Capacitance of a series combination is given by:

$C_{s}=\frac{Q}{V}$

Then the charge stored in the series combination is:

$Q=C_{s} V$

Energy stored in the series combination is:

$U_{c}=\frac{1}{2} V^{2} C_{s}$

Solve the Problem: 

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

b)

Identify the unknown:

The charge and energy stored if the capacitors are connected in parallel

Set Up the Problem: 

Capacitance of a parallel combination:

$C_{p} =C_{1} +C_{2} +C_{3}$

$C_{p} =2+7.4=9.4*10^-6F$

Capacitance of a parallel combination is given by

$C_{p} =\frac{Q}{V}$

Then the charge stored in the parallel combination is

$Q=C_{p} V$

Energy stored in the parallel combination is:

$U_{c}=\frac{1}{2} V^2C_{p}$

Solve the Problem:

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

5 0

3 years ago

Read 2 more answers

A fuel cell vehicle draws 50 kW of power at 70 mph and is 40% efficient at rated power. You are asked to size the fuel cell syst

svetoff [14.1K]

Answer:

a) fuel cell weight and volume: 50 kg, 33.3 L

b) fuel tank weight and volume: 321 kg, 643 L

Explanation:

Fuel Cell

Delivery of 50 kW from a source with a power density of 1.5 kW/L requires a source that has a volume of ...

(50 kW)/(1.5 kW/L) = 33.3 L

The weight of the power source is 1 kW/kg, so will be ...

(50 kW)/(1 kW/kg) = 50 kg

__

Fuel Tank

400 miles at 70 mph will take (400/70) h ≈ 5.71429 h. In that time, the energy used by the vehicle power plant is ...

(50 kW)(5.71429 h)(3600 s/h) = 1028.57 MJ

Since the power plant is 40% efficient, it must be supplied with 2.5 times that amount of energy, or 2571.43 MJ.

The tank volume and mass will then be ...

volume = 2571.43 MJ/(4 MJ/L) = 642.9 L

mass = 2571.43 MJ/(8 MJ/L) = 321.4 kg

_____

Comment on fuel tank volume

It appears the fuel tank would need to be equivalent in size to a sphere about 1.1 m in diameter.

4 0

3 years ago

I need solution for this question please

AlladinOne [14]

Answer:

a true b false thos is the solution

6 0

3 years ago

Read 2 more answers

Set oWMP = CreateObject("WMPlayer.OCX.7?) Set colCDROMs = oWMP.CdromCollection do if colCDROMs.Count >= 1 then For i = 0 to c

vladimir2022 [97]

Answer:

gg

Explanation:

gg

7 0

3 years ago

I have a stream with three components, A, B, and C, coming from another process. The stream is 50 % A, and the balance is equal

tigry1 [53]

Answer:

$X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}$

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = 0.533

Explanation:

A + 2B + 4C ⇒ 2X + 3Y

Given a stream containing 50% A, 25% B and 25% C, to get the limiting reactant, lets take a simple basis

Say stream is 10 moles, this give

A = 5moles

B = 2.5mole

C = 2.5moles

from the balanced equation above,

1mole of A ⇒ 4moles of C

∴ 5moles of A ⇒ (5x4)/1 ⇒ 20moles of C

also;

2mole of B ⇒ 4moles of C

∴ 2.5moles of B ⇒ (2.5x4)/2 ⇒ 5moles of C

so clearly from above reactant C is the limiting reactant.

Note: To get conversion of a process, we must use the limiting reactant. this is because ones it is used up, the reaction comes to an end

Formula to obtain conversion is:

Conversion = (Amount of A used up)/(Amount of A fed into the system)

$X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}$

where, Nₐ₀-Nₐ = is the amount in moles of A used up

Nₐ₀ = amount in moles of A fed into the system

The next question is what mole of reactant C will give 0.1mole fraction of Y

Recall our basis = 10moles

from conservation of mass law, 10mole of product must come out which 0.1 moles fraction is Y

therefore amount Y in the product is = 0.1x10 = 1mole

if 3moles of Y ⇒ 4mole of C

∴ 1mole of Y ⇒ (1x4)/3 ⇒ 1.33moles of C

calculating the conversion of limiting reactant C that will give 0.1mole fraction of Y

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = 0.533

5 0

3 years ago