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strojnjashka [21]
3 years ago
6

When does carbon dioxide absorb the most heat energy? uring freezing during deposition during sublimation during condensation?

Physics
2 answers:
prisoha [69]3 years ago
4 0

Answer;

-During sublimation

Explanation;

-Carbon dioxide in its solid form is known as “dry ice,” because under common conditions it sublimates, turning directly into gas, rather than melting into liquid.

-The enthalpy of sublimation, or heat of sublimation, is the heat required to change one mole of a substance from solid state to gaseous state at a given combination of temperature and pressure,

-The low temperature and direct sublimation to a gas makes dry ice an effective coolant, since it is colder than water ice and leaves no residue as it changes state. Its enthalpy of sublimation is 571 kJ/kg


ololo11 [35]3 years ago
3 0
The answer for the given question above would be the third option. Carbon dioxide absorbs the most heat energy during SUBLIMATION. By definition, sublimation is <span>the transition of a substance from the solid to the gas phase without passing through the intermediate liquid phase. Hope this answers your question.</span>
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A 30 lb weight is attached to the end of a spring. The spring is stretched 6 in. Find the equation of motion if the weight is released from rest a point 3 inches above equilibrium position 。x(,) =-2 sin(81) 32 x(t) =-32 cos(80 O x(r) =-icos(81)

Explanation:

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We intend to observe two distant equal brightness stars whose angular separation is 50.0 × 10-7 rad. Assuming a mean wavelength
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Answer:

13.4cm

Explanation:

According to Rayleigh’s criterion the angular resolution to distinguish two objects is given by:

\theta=1.22\frac{\lambda}{b}

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λ: wavelength of the light = 550nm

b = diameter of the objective

By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:

b=1.22\frac{\lambda}{\theta}=1.22\frac{550*10^{-9}m}{50.0*10^{-7}rad}=0.134m=13.4cm

hence, the smallest diameter objective lens is 13.4cm

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The proper time between two events is measured by clocks at rest in a reference frame in which the two events: The proper time b
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a - As long as the time between 2 events is reconcilable with a light signal, the time between the events, in that frame, can be determined.

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Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is
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Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

W_b = m_bg =  \\ \\  W_b = (0.730 \  kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N

The mass of the water accumulated in the bucket after 3.20s is:

m_w= (0.20 \ L/s) ( 3.20)s

m _w=0.64 \ kg

To determine the weight of the water accumulated in the bucket, we have:

W_w = m_w g

W_w = ( 0.64  \ kg )(9.80\  m  \  /s^2)

W_w = 6.272 \ N

For the speed of the water before hitting the bucket; we have:

v = \sqrt{2gh}

v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

F = v ( \dfrac {dm}{dt} )

F = (8.4 \ m/s)( 0.200 \ L/s)

F = 1.68 N

Finally, the reading scale is:

F_{scale = 7.154 N + 6.272 N + 1.68 N

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