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fomenos
3 years ago
7

Determine the boundary work done by a gas during an expansion process if the pressure and volume values at various states are me

asured to be 300 kPa, 1 L; 290 kPa, 1.1 L; 270 kPa, 1.2 L; 250 kPa, 1.4 L; 220 kPa, 1.7 L; and 200 kPa, 2 L.
Physics
1 answer:
astraxan [27]3 years ago
7 0

Answer:

W=0.217kJ

Explanation:

#Boundary Work is energy expended when a force acts through a displacement

The boundary of work can be calculated as the area under the P-V curve for all the 4 stages measured:

W= \sum_{n=1}^{5} \frac{P_{n+1}+P_{n+2}+P_{n+3}+P_{n+4}+P_{n+5}}{2}\\\\=0.5((290+300)(1.1-1)+(270+290)(1.2-1.1)+(250+270)(1.4-1.3)+(220+250)(1.7-1.4)+(200+220)(2-1.7))\times 10^-^3kJ\\\\=0.217kJ

Hence the boundary of work done by a gas during this 4-stage process is 0.217kJ

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A horizontal spring has spring constant k = 360 N/m. First, compress the spring from its uncompressed length (x = 0) to x = 11.0
mixas84 [53]

Here we can say that by energy conservation principle

Elastic potential energy of spring will convert into kinetic energy of the block

so here we will have

\frac{1}{2} kx^2 = \frac{1}{2}mv^2

we also know that

k = 360 N/m

x = 11 cm

m = 1.85 kg

now we will use all in above equation

\frac{1}{2}\times 360\times (0.11)^2 = \frac{1}{2}\times (1.85) v^2

4.356 = 0.925 v^2

v^2 = 4.71

v = 2.17 m/s

so it will move with speed 2.17 m/s after separating from spring

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3 years ago
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Answer:

C. remix

Explanation:

A is wrong because the prefix ir- means not

B is wrong because the prefix un means not

C is correct because re means again

D is wrong because over means too much

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3 years ago
in 10 minutes a heart can beat 700 times at this rate in how many minutes will a heart beat 140 times at what rate can a heart b
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The table below shows the distance d(t) in feet that an object travels in t seconds:
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The rate of change of d(t) at t = 2 and t = 6 is the ratio between the change of distance (difference between the distances) to the time elapsed. That is,
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4 years ago
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A student on a skateboard is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass
saw5 [17]

Answer:

W = 609.97J

Explanation:

In order to calculate the work done by the student, you take into account that the total work is equal to the change of the kinetic energy of the student, as follow:

W_T=\Delta K\\\\W+W_g-W_f=\frac{1}{2}m(v^2-v_o^2)\\\\W+Mgsin\alpha d-F_fd=\frac{1}{2}m(v^2-v_o^2)         (1)

The work done by the friction force is negative because it is against the motion of the student.

W: work done by the student = ?

Wf: work done by the friction force

Wg: work done by the gravitational force

Ff: total friction force = 41.0N

m: mass of the skateboard = 53.0kg

d: distance traveled by the student

v: final speed of the student = 6.90m/s

vo: initial speed of the student = 1.40m/s

α: angle of the incline

You first calculate the distance d, with the Pythagoras' theorem

d=\sqrt{(2.15m)^2+(12.4m)^2}=12.58m

Furthermore, the angle α is:

\alpha=tan^{-1}(\frac{2.15m}{12.4m})=9.83\°

Then, you solve the equation (1) for W and replace the values of all parameters:

W=\frac{1}{2}m(v^2-v_o^2)+F_fd-Mgsin\alpha d\\\\W=\frac{1}{2}(53.0kg)((6.90m/s)^2-(1.40m/s)^2)+(41.0N)(12.58m)\\-(53.0kg)(9.8m/s^2)sin(9.83\°)(12.58m)\\\\W=609.97J

The work done by the student is 609.97J

6 0
4 years ago
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