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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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2 years ago

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77

Vikki [24]

The formula for the rotational kinetic energy is

$KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2}$

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

$I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2} =2.07459 \ kgm^{2}$

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

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3 years ago

A space craft is moving relative to the earth , an observer on the earth finds that, between 1pm and 2pm according to her clock,

madam [21]

The speed of the space craft relative to the earth is given as: 0.024c. This is solved using the the equation for time dilation.

<h3>What is time dilation?</h3>

Time dilation is the "slowing down" of a clock as determined by an observer in relative motion with regard to that clock under the theory of special relativity.

The formula is given as :

Δt = [Δr]/ √ 1 - (v²/c²)

Thus,

v = c√1 - (Δr/Δt)²

= c √(1 - (3600/3601)²

v = 0.024c

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brainly.com/question/1933572
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1 year ago

A grocery cart with a mass of 15 kg is being pushed at constant speed up a 12∘ ramp by a force FP which acts at an angle of 17∘

Alenkasestr [34]

Answer: a. 198.6J b. - 198.6J

Explanation: Parameters given:

m = 15kg

g = 9.8m/s²

∅ = 12°

a. Work done by the force Fp on the cart if the ramp is 6.5m long.

Given the formula, Fp = Mgsin∅ = 15 x 9.8 x sin12° = 30.56N

Therefore Work done (Wp) = Fp x Ramp Length = 30.56 x 6.5 = 198.64Nm or 198.6J

b. The work done by the force mg on the cart.

Since the cart is being pushed upwards, it acts against gravity with its direction of motion. Taking into account the formula from the previous answer for Work Done (Wg) = Fmg x distance

= 15kg x -9.8m/s² x Sin12° x 6.5m

= - 198.6J

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