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nlexa [21]
4 years ago
7

What happens to the size of the static frictonal force as you begin to pull on the force scale

Physics
2 answers:
LiRa [457]4 years ago
4 0
Not sure they are asking this, but it may well be:

Static frictional force can be from 0 to a maximum value. For instance, a book on your desk does not move, but the frictional force is zero. Then, when you try to move it, it doesn't until a moment where it does. The maximum static force is a coefficient (mu_static) multiplied by the Normal force (that in turn is usually derived from the weight, but whichever it is).

Then, after set in motion, static friction does not apply. It is kinetic frictional force, whose formula is the same, but it is just the value (not between zero and a maximum value):

F_kinetic = mu_kinetic * Normal

What happens is that mu_kinetic < mu_static and the frictional force will be less!

Once in motion, it's easy to keep it moving, right? Hope it helps!
Veronika [31]4 years ago
3 0
It is at its peak force, it takes the most force to overcome the static stare because of inertia
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To overcome an object's inertia, it must be acted upon by __________.
Nastasia [14]

Correct answer choice is :


C) Force


Explanation:


In physics, a force is any cooperation that, when unrestricted, will vary the motion of an object. A force can create an object with mass to alter its velocity to accelerate. Force can also be defined intuitively as a push or a pull. A force acting on an object may create the object to alter shape, to start moving, to stop moving, to stimulate or decelerate. When two objects communicate with each other they exert a force on each other, the forces are just in size but different in direction.

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4 years ago
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Why are some substances gases at room temperature but others are liquid or solid?
Otrada [13]

Answer:

The presence of strong intermolecular forces favors a condensed state of matter. liquid or solid), while very weak intermolecular interaction favor the gaseous state.

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2 years ago
An ideal gas initially at 4.00atm and 350 K is permitted
Nuetrik [128]

Explanation:

It is given that initially pressure of ideal gas is 4.00 atm and its temperature is 350 K. Let us assume that the final pressure is P_{2} and final temperature is T_{2}.

(a)   We know that for a monoatomic gas, value of \gamma is \frac{5}{3}[/tex].

And, in case of adiabatic process,

                PV^{\gamma} = constant              

also,         PV = nRT

So, here    T_{1} = 350 K,    V_{1} = V,  and   V_{2} = 1.5 V

Hence,      \frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}

         \frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{5}{3} -1}

          T_{2} = 267 K

Also,   P_{1} = 4.0 atm,   V_{1} = V,  and   V_{2} = 1.5 V

        \frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}

        \frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{5}{3}}

            P_{2} = 2.04 atm

Hence, for monoatomic gas final pressure is 2.04 atm and final temperature is 267 K.

(b) For diatomic gas, value of \gamma is \frac{7}{5}[/tex].

As,        PV^{\gamma} = constant              

also,         PV = nRT

T_{1} = 350 K,    V_{1} = V,  and   V_{2} = 1.5 V

              \frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}

         \frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{7}{5} -1}

          T_{2} = 289 K

And,   P_{1} = 4.0 atm,   V_{1} = V,  and   V_{2} = 1.5 V

                \frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}

        \frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{7}{5}}

            P_{2} = 2.27 atm

Hence, for diatomic gas final pressure is 2.27 atm and final temperature is 289 K.

6 0
3 years ago
Can some please help me
Brilliant_brown [7]

Answer:

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Explanation:

https://web.cortland.edu/buckenmeyerp/Lecture14.html

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