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Mkey [24]
4 years ago
10

which do you think would be warmer on a winter day when there is no wind, a thick forest or a grassy field

Physics
2 answers:
Alex Ar [27]4 years ago
5 0
It depends on the location but I believe the grassy field since grass grows by photosynthesis so there would be a lot on sunlight so there would be alot of warmth
nadya68 [22]4 years ago
3 0
It depends on where you are at some placers might have a coldness which thick forest and some may no. It al. Depends on your location and where you are at at that moment
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The infant's tendency to turn its head toward things that touch its cheek is known as the
katovenus [111]

Answer:

I think it is <em><u>Rooting</u></em><em> </em><u><em>Reflex</em></u>

4 0
4 years ago
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How does increasing the distance between charged objects affect the electric force between them? the electric force increases be
konstantin123 [22]

the electric force decreases because the distance has an indirect relationship to the force

Explanation:

The electric force between two objects is given by

F=k \frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the distance between the two objects

As we can see from the formula, the magnitude of the force is inversely proportional to the square of the distance: so, when the distance between the object increases, the magnitude of the force decreases.

3 0
3 years ago
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A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1
balandron [24]

Answer:

The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

Explanation:

For point a:

Energy balance equation:

\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

W=0\\\\Q=0\\\\m_e=0

From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

temperature:

T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

 \frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\

For point c:

 Entropy is given by the following formula:

\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}

5 0
3 years ago
Why does gravitational energy increase away from earth
aliina [53]
Because of the rule of mask.
4 0
3 years ago
Ice has a specific heat of 2.093Jg ∘C, liquid water's specific heat is 4.184Jg ∘C, and steam has a specific heat of 1.864Jg ∘C.
Ivan

Answer:

The liquid phase will have the lowest temperature change upon heating.

Explanation:

Assuming no phase change due to heating, we know that the temperature change, is proportional to the mass heated, being the proportionality constant  a quantity that depends on the material, and represents the resistance of the material to change the temperature, called specific heat.

So, if we assume that the mass is the same for the three phases, and that the amount of heat supplied is also the same,the phase with the highest specific heat will have the lowest temperature change.

So, the liquid phase will be the one that exhibits this behavior, as the specific heat of liquid water (4.184 J/gºC) is the highest among the three phases.

8 0
4 years ago
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