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3 years ago

20-point physics question! :D

2 answers:

Alex3 years ago

8 0

Your hands get warm by a fire because chemical energy is converted into heat energy.

The wood - has chemical energy when you burn it, and then, heat waves rise to the air, which means that we went from :

Chemical energy ⇒ Heat energy

Alright, now, also think about, that energy can never be created not destroyed, so it was obviously transferred, and yet conserved.

If this doesn't make sense, please message me, or comment below. :))

Virty [35]3 years ago

6 0

I think, it must be a chemical. Because the fire is a chemical reaction. So, the chemical energy converted to heat energy.

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Suppose two comets, comet A and comet B, were orbiting the sun, having the same average orbital radii. If comet A had a higher o

zhannawk [14.2K]

Answer:

a. A

Explanation:

Kepler's First Law says that the orbits of planets are ellipses with the sun at one focus of the ellipse. Moreover, Kepler’s Second Law says that the line joining the planet to the sun sweeps out equal areas in equal times as it moves along its orbit. Finally, Kepler’s Third Law says that the ratio of the squares of the periods for two planets is equal to the ratio of the cubes of their semi-major axes.

By these laws, the comet A will have lower orbital speed.

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3 years ago

In a real hemoglobin molecule, the tendency of oxygen to bind to a heme site increases as the other three heme sites become occu

sasho [114]

Since there are four states, then the grand partition function of the system is

$Z = 1+2 e^{(0.55eV+ \alpha )/kT} +e^{(1.3eV+ 2\alpha )/kT}$

where α is the chemical potential

Then, the occupancy of the system is

$bar \ n=(e^{(0.55eV+ \alpha )/kT} +e^{(1.3eV+ 2\alpha )/kT})/Z$

Then using this equation, $\alpha =-kT(V Z_{int} /N v_{Q})$ and approximating Z_int to be kT/0.00018 eV, the model would look as that attached in the figure. That is the occupancy vs. pressure graph.

There are more occupancies when the oxygen is high (high pressure) especially in the lungs. Heme sites tend to be occupied by oxygen.

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3 years ago

Lake Baikal in Siberia has a maximum depth of 1642 m. What is the water

Veseljchak [2.6K]

Explanation:

Fluid gauge pressure is:

P = ρgh

where ρ is the fluid density and h is the depth of the fluid.

P = (1000 kg/m³) (9.8 m/s²) (1642 m)

P = 16,091,600 Pa

Rounded to four significant figures, the gauge pressure is 16.09 MPa.

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3 years ago

A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow

Ksivusya [100]

Answer:

The answer is 1.87nm/s.

Explanation:

The $110g/hr$ water loss must be replaced by $110g/hr$ of sap. 110g of sap corresponds to a volume of

$110g \div \dfrac{1040*10^3g}{1*10^6cm^3} = 106cm^3$

thus rate of sap replacement is

$106cm^3/hr = 106*10^{-6}m^3/3600s = 2.94*10^{-8}m^3/s$

The volume of sap in the vessel of length $x$ is

$V = Ax$ ,

where $A$ is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

$V = 2000Ax$

taking the derivative of both sides we get:

$\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}$

on the left-hand-side $\dfrac{dx}{dt}$ is the velocity $v$ of the sap, and on right-hand-side $\dfrac{dV}{dt} = 2.94*10^{-8}m^3/s$ ; therefore,

$2.94*10^{-8}m^3/s=2000Av$

and since the cross-sectional area is

$A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2$ ;

therefore,

$2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v$

solving for $v$ we get:

$v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}$

$\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}$

which is the upward speed of the sap in each vessel.

8 0

3 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:

$U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )$