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Sedaia [141]
3 years ago
15

A 0.40-μF capacitor is connected to a 5.0-V battery. How much charge is on each plate of the capacitor? Express your answer usin

g two significant figures.
Physics
1 answer:
weqwewe [10]3 years ago
7 0

Answer:

Q=2.0\times10^{-6}C

Explanation:

The equation of charge on the capacitor given its capacitance and a voltage applied to it is Q=CV.

Our capacitance is expressed in <em>microfarads</em>, where micro stands for \times10^{-6}, so in S.I. we sill have:

Q=CV=(0.4\times10^{-6}F)(5V)=0.000002C=2.0\times10^{-6}C

Where the results has been expresed with <em>two significant figures</em>.

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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

<u>At State 1:</u>

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

<u>At State 1:</u>

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_aP_{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.

<u>At State 1;</u>

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

<u>At State 2:</u>

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 yo-yo has a string that is 0.95 m in length. What is the period of oscillation if the yo-yo is allowed to swing back and forth
yulyashka [42]

Answer:

Explanation:

As we know the , equation of time period for simple pendulum ,

T = 2*pi*\sqrt{l/g}

hence putting values we get ,

the solution is in picture ,

please  

Brain-list it or support me at my U-Tube channel " ZK SOFT&GAMING " I will be thankful

5 0
3 years ago
The density of gold is 19.3. G/cm(3). If a nugget of iron pyrite and nugget of gold each have a mass of 50 g, what can you concl
tangare [24]
The pyrite will be bigger, because its density is much lower.

I <em>do</em> know that the gold's volume will be 2.5906 (With a bunch more numbers after it)

50 divided by 19.3 = 2.5906
5 0
3 years ago
Explain in detail what information can be learn about atoms of different elements by examining the periodic table . For the Maxi
raketka [301]

Answer:

We know the information about atomic size, energy, electronic configuration etc. of atom from the periodic table.    

Explanation:

  • Periodic table is the arrangement of elements that are arranged according to their properties and electronic configuration.
  • In periodic table, on furthest right side of the periodic table, noble gases like He, Ne, Ar etc are arranged.
  • The atomic number of element increases while moving from left towards right in the periodic table.
  • The metallic character of element decreases as we proceed the table towards right.
  • They readily accept electron to fill the valence shell hence becoming more metallic in character.  
5 0
3 years ago
If velocity is decreasing,then acceleration:
ziro4ka [17]
If velocity is decreasing, then acceleration is in the direction
opposite to the velocity.

If the object is moving in the direction that you call 'positive',
then acceleration is negative.
5 0
3 years ago
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