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Lapatulllka [165]
2 years ago
7

A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte

rnal energy of this gas. The value of the gas constant is 8.31451 J/mol * K. Answer in units of J.
Physics
1 answer:
Vitek1552 [10]2 years ago
5 0

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

     = 3/2×(5.3)×(8.31451)×(24 + 273)

     = 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

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A crate is lifted vertically 1.5 m and then held at rest. The crate has weight 100 N (i.e., it is reese (enr647) – HS OnRamps 04
icang [17]

Answer:

W = 0 J

Explanation:

It is given that,

Weight of the crate, W = 100 N

Distance moved by the crate, d = 1.5 m

Let W is the work done to hold the crate 1.5 m above the ground in this way. It is given by the product of force and the displacement. Its formula is given by :

W=F\times d\times cos\theta

Here, \theta=90^{\circ} as it is lifted vertically

W=100\ N\times 1.5\ m\ cos(90)      

W = 0

So, the work done to hold the crate is 0. Hence, this is the required solution.

5 0
2 years ago
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A large mass collides with a stationary, smaller mass. How will the masses behave if the collision is inelastic?
Mamont248 [21]
Most likely they would stick together and keep moving together
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2 years ago
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Which of the following exists around every object that has mass?
ICE Princess25 [194]

Answer:

A is the correct answer.

Explanation:

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2 years ago
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A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa
elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0
2 years ago
In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into t
Pie

Answer:

A) Q_a=74256\ J

B) Q=93562560\ J

Explanation:

Given:

  • temperature of air, T_a=-19+273=254\ K
  • temperature of lungs, T_l=37+273=310\ K
  • specific Heat exchanged from the lungs , c_l=0.47\ J.kg^{-1}.K^{-1}
  • specific heat of air, c_a=1020\ J.kg^{-1}.K^{-1}
  • mass of 1 L air, m'=1.3\ kg
  • breath rate, b=21\ breath.min^{-1}

A)

Now,

amount of heat needed to warm the air of lungs to the body temperature:

Q_a=m'.c_a.\Delta T

Q_a=1.3\times1020\times (310-254)

Q_a=74256\ J

B)

Amount of heat lost per hour:

<u>No. of breaths per hour:</u>

B=b.60

B=21\times 60

B=1260

<u>Now the total loss of energy in 1 hr.:</u>

Q=Q_a.B

Q=74256\times 1260

Q=93562560\ J

7 0
2 years ago
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