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Ganezh [65]
3 years ago
14

A sample of n2 gas occupies a volume of 746 ml at stp. What volume would n2 gas occupy at 155 ◦c at a pressure of 368 torr?

Physics
1 answer:
musickatia [10]3 years ago
7 0

Answer:

2.41 L

Explanation:

We can solve the problem by using the ideal gas equation, which can be rewritten as:

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

where we have:

p_1 = 1.01\cdot 10^5 Pa (initial pressure is stp pressure)

V_1 = 746 mL = 0.746 L = 7.46\cdot 10^{-4}m^3 is the initial volume

T_1 = 0^{\circ}=273 K is the initial temperature (stp temperature)

p_2 = 368 torr = 4.9\cdot 10^4 Pa is the final pressure

V_2 = ? is the final volume

T=155^{\circ}=428 K is the final temperature

By substituting the numbers inside the formula and solving for V2, we find the final volume:

V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}=\frac{(1.01\cdot 10^5 Pa)(7.46\cdot 10^{-4} m^3)(428 K)}{(273 K)(4.9\cdot 10^4 Pa)}=2.41\cdot 10^{-3} m^3

which corresponds to 2.41 L.

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ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

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                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

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3 years ago
Kate gathered 3 of the same size made of different materials: glas, clear plastic, and aluminum painted black. She placed on the
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<u>Answer:</u>

<em>To relate the type of box material to the warmth inside the box </em>

<u>Explanation:</u>

3 boxes are made with three different materials glass, plastic and Aluminium. Thermal conductivity is different for different materials. thermal conductors allow easy flow of heat through them and insulators allow minimal or no flow of heat through them.

Thus the amount by which an object gets heated up depends on the value of its conductivity. In this experiment, glass and plastic are insulators and aluminium is a conductor. Among glass and plastic, plastic is a better insulator.

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3 years ago
Please need help fast
iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) 1.5 m/s^2

The average acceleration of the horse can be calculated as:

a=\frac{v-u}{t}

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v is the final velocity

u is the initial velocity

t is the time interval

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u = 0

v = 15 m/s

t = 10 s

Substituting,

a=\frac{15-0}{10}=1.5 m/s^2

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For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

s=ut+\frac{1}{2}at^2

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s is the distance travelled

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t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

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Substituting,

s=0+\frac{1}{2}(1.5)(10)^2=75 m

(d) See attached graphs

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x=ut+\frac{1}{2}at^2

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a=1.5 m/s^2 (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

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Answer:A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no ..

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2 years ago
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