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Julli [10]
3 years ago
15

A certain gas is present in a 15.0 l cylinder at 2.0 atm pressure. if the pressure is increased to 4.0 atm the volume of the gas

decreases to 7.5 l . find the two constants ki, the initial value of k, and kf, the final value of k, to verify whether the gas obeys boyle's law by entering the numerical values for ki and kf in the space provided.
Physics
1 answer:
raketka [301]3 years ago
4 0

In the first condition: P1 = 2.0 atm V1 = 15.0 L P1V1 = Ki 2 * 15 = Ki Ki = 30

In the second condition Pt = 4.0 atm Vt = 7.5L Pt * Vt = Kt 4.0 * 7.5 = Kt Kt = 30

Conclusion: Ki = Kt. Since the two values of k are the same. Therefore, the gas obeys Boyle's Law. Take note Boyle’s law is an ideal gas law which tells us that the volume of an ideal gas is contrariwise proportional to its absolute or total pressure at a constant temperature.
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Answer:

It would not be possible the cohesion among water molecules by the polar covalent bonding.

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A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

-W_{x}+f_{s}=0

when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

When seeing the free-body diagram we can determine that:

W_{y}=mg cos \alpha

so we can substitute that in the sum of y-forces equation, so we get:

N=mg cos \alpha

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

mg cos \alpha \mu_{s}=mg sin \alpha

we can divide both sides of the equation into mg so we get:

cos \alpha \mu_{s}=sin \alpha

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\mu_{s}=\frac{sin \alpha}{cos \alpha}

when simplifying this we get that:

\mu_{s}=tan \alpha

now we can solve for the angle so we get:

\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

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α=19.29°

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α=19°

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