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Dahasolnce [82]
3 years ago
7

Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note

that the latent heat is equal to the change in enthalpy, and that liquid nitrogen boils at 77 K. The system is kept at a constant pressure of 1 atm. Assuming that you can treat the gas as ideal gas and that the volume of the liquid is approximately zero, compute the binding energy of a nitrogen molecule in the liquid. (the binding energy is the difference in internal energy per molecule between the liquid and gas)
a. 4.2e^-18 J
b. 9.4e^-21 J
c. 4.8e^ 24 J
d. 2.1e^-19 J
e. 1.3e^-23 J
Physics
1 answer:
Anastaziya [24]3 years ago
7 0
Suppose that we use a future boil liquid nitrogen
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Answer: question 1 , would be one question 2 , would be 1 joule and number three would be number one and number four would be , power and last one would be, number two

Explanation: sorry if its wrong

5 0
3 years ago
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
3 years ago
A computational model predicts the maximum kinetic energy a roller coaster car can have given its mass, the speed at the highest
Ymorist [56]

Answer:

D

Explanation:

The decrease in potential energy is equal to the increase in kinetic energy.

mgh

250 x 9.8 x 30

=73, 500

3 0
3 years ago
An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50
Setler79 [48]

Answer:

very hard others will answer it

Explanation:

hard

6 0
3 years ago
A sample of helium (He) occupies 8.0 liters at 1 atm and 20.0◦C. What pressure is necessary to change the volume to 1.0 liters a
nevsk [136]

Apply the combined gas law

PV/T = const.

P = pressure, V = volume, T = temperature, PV/T must stay constant.

Initial PVT values:

P = 1atm, V = 8.0L, T = 20.0°C = 293.15K

Final PVT values:

P = ?, V = 1.0L, T = 10.0°C = 283.15K

Set the PV/T expression for the initial and final PVT values equal to each other and solve for the final P:

1(8.0)/293.15 = P(1.0)/283.15

P = 7.7atm

7 0
3 years ago
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