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lesya692 [45]
1 year ago
8

Which process requires more energy: completelyvaporizing 1 kg of saturated liquid water at 1 atm pressure orcompletely vaporizin

g 1 kg of saturated liquid water at 8 atmpressure?
Physics
1 answer:
goldfiish [28.3K]1 year ago
5 0

The correct answer would be the first option. That means completely vaporizing 1 kg of saturated liquid water at 1 atm pressure would require more energy.

<h3>Enthalpy of Vaporization</h3>

The enthalpy of vaporization, also known as the heat of vaporization or heat of evaporation, is the quantity of energy that must be added to a liquid substance in order to transform a quantity of that substance into a gas.

The enthalpy of vaporization decreases as the pressure increases, so it takes more energy to completely vaporize 1 kg of saturated liquid water at 1 atm than at 8 atm.

<h3>What is the enthalpy of the vaporization of water? </h3>

The enthalpy of vaporization depends on the pressure at which that transformation occurs.

40.65 kJ/mol

The heat of vaporization value for water is 40.65 kJ/mol.

To know more about Vaporization enthalpy visit:

brainly.com/question/7311854

#SPJ4

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Current always returns to the negative terminal.

<u>Explanation</u>:

Current flows from a positive to a negative terminal. It is due to the flow of negative electrons to positive side. The current flows from a high potential to a low potential. There is high concentration of electrons at the positive terminal and low concentration at the negative terminal. When the circuit is complete the electrons move from a high concentration to low concentration i.e; from positive terminal to negative terminal.

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One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the f
Andreas93 [3]

Answer:

Part a)

\theta_2 = 15 degree

Part b)

\Delta t = 2.88 s

Explanation:

Part a)

In order to have same range for same initial speed we can say

R_1 = R_2

\frac{v^2 sin2\theta_1}{g} = \frac{v^2 sin2\theta_2}{g}

so after comparing above we will have

\theta_1 = 90 - \theta

so we have

75 = 90 - \theta_2

\theta_2 = 15 degree

Part b)

Time of flight for the first ball is given as

T_1 = \frac{2vsin\theta}{g}

T_1 = \frac{2(20)sin75}{9.81}

T_1 = 3.94 s

Now for other angle of projection time is given as

T_2 = \frac{2(20)sin15}{9.81}

T_2 = 1.05 s

So here the time lag between two is given as

\Delta t = T_1 - T_2

\Delta t = 3.94 - 1.05

\Delta t = 2.88 s

5 0
3 years ago
On a summer day, Thomas is standing in his driveway waiting for his mother. Energy from the Sun is making Thomas feel hot. This
Leno4ka [110]

Answer:

A. Energy streaming through space toward Thomas

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4 0
3 years ago
7. A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a level (frictionless) surfa
makkiz [27]

Answer:

0.505 m

Explanation:

From the question,

The kinetic energy of the car = energy stored in the spring

1/2mv² = 1/2ke²...................... Equation 1

Where m = mass of the car, v = velocity of the car, k = spring constant of the car, e = extension/compression

make e the subject of the equation

e = v√(m/k)............... Equation 2

We can calculate the value of v, by applying,

v² = u²+2gH...................... Equation 3

Where u = initial velocity of the car, H = height of the car, g = acceleration due to gravity.

Given: u = 0 m/s (from rest), H, 10 m, g = 9.8 m/s²

Substitute into equation 2

v² = 2(10×9.8)

v² = 196

v = √196

v = 14 m/s

Also given: m = 1300 kg, e = 1.0×10⁶ N/m =1000000 N/m

Substitute into equation 2

e = 14√(1300/1000000)

e = 14√(0.00013)

e = 14(0.036)

e = 0.505 m

Hence the maximum distance of the spring is compressed = 0.505 m

5 0
3 years ago
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