All categories

4 years ago

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

2 answers:

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

You might be interested in

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 Rooting Reflex

4 0

4 years ago

Read 2 more answers

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

Read 2 more answers

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