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makvit [3.9K]
4 years ago
12

Five moles of an ideal monatomic gas with an initial temperature of 127∘C expand and, in the process, absorb 1500 J of heat and

do 2100 J of work. What is the final temperature of the gas?
Physics
1 answer:
Alja [10]4 years ago
8 0

Answer:

117.38^{\circ}C

Explanation:

We are given that

Initial temperature=T_1=127^{\circ}

Number of moles=n=5

Heat absorbed,Q=1500 J

Work done=W=2100 J

We have to find the final temperature  of the gas.

\Delta U=Q-W=1500-2100=-600 J

T_2=\frac{2\Delta U}{3nR}+T_1

Where R=8.314 J/mol-K

Using the formula

T_2=\frac{2\times (-600)}{3\times 5\times 8.314}+127^{\circ} C

T_2=117.38^{\circ}C

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NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

A)

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

t=\frac{2u sin \theta}{g}

where

u is the initial speed

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g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is t.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t

B)

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

h=\frac{u^2 sin^2\theta}{2g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is h.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h

C)

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

D=\frac{u^2 sin(2\theta)}{g}

where:

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D

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3 years ago
What is the simplest way to build a graph of kinetic and potential energy?
Elena-2011 [213]

Answer:

analize the levels of kinetic and potential energy and look up a guide to graph it and follow that

Explanation:

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3 years ago
How old is the universe
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Roughly 13.8 billion years old according to science
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The acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
VladimirAG [237]

We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

  • it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force  is mathematically given as

F=\frac{F}{dx}

Therefore

F=-kdx

k=600Nm^{-1}

now

K.E=0.5x ds^2

K.E=600*(-0.1^2)

K.E=3J

Therefore

the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

For more information on this visit

brainly.com/question/23379286

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3 years ago
Awave has a period of 1. What is its frequency?
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Answer:

Frequency, f = 1 unit

Explanation:

It is given that,

Period of the wave, T = 1 unit

We need to find the frequency of the wave. There exist an inverse relationship between period and the frequency of the wave. It is given by :

T=\dfrac{1}{f}

Or

f=\dfrac{1}{T}

f=\dfrac{1}{1\ units}

f = 1 unit

So, the frequency of the wave is 1 unit. Hence, this is the required solution.

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