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charle [14.2K]
3 years ago
9

We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 129 ∘C. The gas expands and, in the process, a

bsorbs an amount of heat equal to 1180 J and does an amount of work equal to 2160 J . What is the final temperature Tfinal of the gas? Use R = 8.3145 J/(mol⋅K) for the ideal gas constant.

Physics
1 answer:
harkovskaia [24]3 years ago
6 0

The solution is in the attachment

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What are noble gases?
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A spaceship hovering over the surface of Venus drops an object from a height of 17 m. How much longer does it take to reach the
Paraphin [41]

1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion  that relates the height as follow:

h=v_{0} t+\frac{gt^{2}}{2}

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

h=\frac{gt^{2}}{2}

We know the height h of the spaceship hovering, and the gravity of Venus is g=8.87\frac{m}{s^{2}}. Substituting this values in the equation h=\frac{gt^{2}}{2}:

17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth g=9.81\frac{m}{s^{2}}.

t=\sqrt{\frac{2(17m)}{9.81\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{9.81\frac{m}{s^{2} } } }=1.86s

8 0
3 years ago
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What would the final temperature be if 8.94 X 10 3 joules of heat were added to 454 grams of copper at 23.0 o C?
ozzi

Answer: 74.1^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=8.94\times 10^3 Joules

m= mass of copper = 454 g

c = specific heat capacity = 0.385J/g^0C

Initial temperature of the copper = T_i = 23.0°C

Final temperature of the water = T_f  = ?

Change in temperature ,\Delta T=T_f-T_i

Putting in the values, we get:

8.94\times 10^3=454\times 0.385\times (T_f-23.0)^0C

T_f=74.1^0C

The final temperature of copper will be 74.1^0C

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Simplify log√8÷log√8​
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Answer:

The answer is 1.

Explanation:

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