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2 years ago

Investigar la función de los espejos esféricos y como funciona, en el microscopio.

Physics

1 answer:

NikAS [45]2 years ago

4 0

Answer:

yea

Explanation:

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in as small room fan of rating 50 watt is used for 10hrs 2bulb of rating 10v are used for 8hrs daily.calculate monthly. electric

Pie

Answer:

The electric bill for June is Rs198000

Explanation:

Convert volt to watt, but in order to do so I need to know the amps and since it is not provided I converted if the amps was 1.

I multiple 50 with 10 then with 30 so I know how much watt the fan takes at June.

Since there are 2 light bulb I multiple 10 with 2 than with 8 than with 30.

15000 watts for the fan,

4800 watts for light bulb,

add them and then times it by 10.

Rs198000

4 0

2 years ago

How can waves travel through ground?

Artyom0805 [142]

The vibration caused by p waves is a volume changes, alternatimg from compression to expansión in the direction that the waves is traveling.

7 0

3 years ago

A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on

svet-max [94.6K]

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}$

Put the value into the formula

$\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}$

$\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}$

$\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}$

$\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}$

$\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1$

$x=0.32\ m$

Hence, The position on the x axis is 0.32 m.

5 0

2 years ago

Which of the following is a true statement?

pishuonlain [190]

I think the answer is A.........

7 0

2 years ago

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