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3 years ago

For 0.37 moles of oxygen (02) gas at room temperature with active translational and rotational degrees of freedom.

Physics

1 answer:

Lelechka [254]3 years ago

6 0

Answer:

B. 161.5 J

Explanation:

$n$ = Number of moles = 0.37

$\Delta T$ = Rise in the temperature of the oxygen gas = 15 K

$Q$ = heat added in order to raise the temperature

$c_{p}$ = specific heat at constant pressure = 3.5

At constant pressure, heat is given as

$Q = n c_{p} R \Delta T\\Q = (3.5) (0.37) (8.314) (15)\\Q = 161.5 J$

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Answer:

33= 0+10 * t so t = 33 / 10 = 3.3 hrs

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The airplane is flying with a constant velocity. Which force acting on the airplane below represents the friction from air resis

tankabanditka [31]

The correct answer to the question is C i.e C represents the friction from air resistance.

EXPLANATION:

Before coming into any conclusion, first we have to understand friction.

The friction is the opposing force which acts tangentially between two bodies in contact when there is a relative motion between them.

The air resistance is that frictional force which is provided by the air to the moving body through it. Hence, the friction from air resistance will be directed opposite to the motion of the body.

In the given diagram, the airplane is going horizontally. The force A acts in forward direction while force C acts in backward direction. The forces B and D are acting vertically. There is no motion in vertical direction. Hence, the net force of A and C will cause the airplane to move.

As the plane is moving along the direction of A, the frictional force must act along the direction of C.

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3 years ago

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How do you know the earth is rotating on its axis

cupoosta [38]

Answer:

Read below!

Explanation:

You can watch the sun wheel across the sky during the day, and the stars at night. Focus a telescope on any star besides the north star--especially southern stars--and you can watch them drift across your field of view.

An alternative explanation is that all the stars are painted on (or holes in) some canopy that rotates around the earth. This explanation does not account for the motion of the "wanderers," or planets, as the Greeks called them, or for the path of the moon among the stars.

As we know the stars are massive bodies of significant and varying distance to the earth, the notion they all swing around us in unison seems highly implausible

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3 years ago

A 1.2kg ball rolls forward with an acceleration of 1.11 m/s. What is the net force on the ball

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Answer:

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1 year ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.