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3 years ago

what does it mean if the coefficient of friction is greater? for example,a coefficient of friction of 0.1 compared to 0.7

Physics

2 answers:

Paul [167]3 years ago

4 0

If the coefficient of a fraction is greater, than the overall fraction is greater than 1. In your example you misspelled fraction wrong, and gave an incorrect example which momentarily confused me, but 0.1/0.7 is a fraction where the numerator/coefficient is smaller than the one on the bottom, or the denominator. So the overall fraction is less than 1.
Hope this helps!

Sergeu [11.5K]3 years ago

3 0

Well, 0.1 is actually less than 0.7, but I understand what you're asking.

The coefficient of friction describes the relationship between two surfaces
that are sliding by each other. The higher the coefficient of friction is, the
'rougher' the meeting is, and the harder it is for one to slide over the other.
A skate blade against ice has a very low coefficient of friction. Sandpaper
against blue jeans has a high coefficient of friction.

A higher coefficient of friction means that when one thing is sliding over
the other one, friction robs more energy from the motion. It's harder to
push one thing over the other one, and when you let go, the moving one
slows down and stops sooner.

Air resistance is actually an example of friction. It prevents falling things
from falling as fast as they would if there were no air. The coefficient of
friction when something moves through air is pretty low. If the same
object were trying to move through molasses or honey, the coefficient
of friction would be greater.

Friction robs energy, and turns it into heat. So, especially in machinery with
moving parts, we want to make the coefficient of friction between the moving parts
as small as possible. That's what the OIL in a car's engine is for.

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At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

$V = V_R + V_L$

The potential difference across the inductance is given by

$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

where

$\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s$ is the time constant of the circuit

At time t=0,

$V_L(0) = V e^0 = V = 20 V$

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

$V_R = V-V_L = 20 V-20 V=0$

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

$V_R = V-V_L = 20 V- 0 = 20 V$

(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

(d) at $t=5.77\cdot 10^{-4}s$

We said that the two voltages will be equal when

$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

$e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s$

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

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Find the final velocity if the initial velocity of 8 m/s with an acceleration of 7 m/s2 over a 3 second interval?

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What are the complementary colors of magenta? Blue? And cyan?

zysi [14]

Answer:

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Explanation:

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pav-90 [236]

if in series one lightbulb burns out the rest are unable to turn on.

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An object that is slowing down in a positive direction must have

zepelin [54]

Answer:

Positive velocity and negative acceleration

Explanation:

An object moving in the positive direction has a positive velocity.

An object that's slowing down while moving in the positive direction has a negative acceleration.

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