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4 years ago

What happens to the coefficient of friction when the weight is increased? Why is this?

Physics

2 answers:

sp2606 [1]4 years ago

8 0

Answer:

The coefficient of friction does not change.

Explanation:

The reason for this is that the coefficient friction depends only on one variable, which is the nature of the surfaces in contact. Therefore, weight has no effect whatsoever on the coefficient of friction. hence, No change occurs in the coefficient of friction if the weight is changed.

hope it helped ya.

Crazy boy [7]4 years ago

6 0

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

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Accuracy in scientific investigation is important because

77julia77 [94]

It ensures good results

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3 years ago

Katie and Mark sit next to one another in class. She has a mass of 40 kg and his mass is 65 kg.

OlgaM077 [116]

A :-) for this question , we should apply
F = GMm by d^2
( For making the calculation easy , first remove the decimals )
Given : G = 6.7 x 10^-11 Nm^2 / kg^2
= 67 x 10^-12 Nm^2 / Kg^2
M = 65 kg
m = 40 kg
d = 0.5 m
Solution -
F = GMm by d^2
F = 67 x 10^-12 x 65 x 40 by 0.5 x 0.5
F = 4355 x 40 x 10^-12 by 0.25
F = 174200 x 10^-12 by 0.25
F = 696800 x 10^-12

.:. The Gravitational force between mark and Katie is 696800 x 10^-12

6 0

3 years ago

How many molecules is in this formula 3H2O

jarptica [38.1K]

Answer: There are 9 molecules.

Explanation: 3H2O has 6 hydrogen atoms and 3 oxygen atoms. Heres a diagram.

HHO + HHO + HHO = 3H2O

7 0

3 years ago

A secant and a tangent meet at a 90Ă‚Â° angle outside the circle. What must be the difference between the measures of the interc

kozerog [31]

The difference between the measures of the intercepted arcs of the given circle is 45°.

A circle is defined as the set of points in a plane equidistant to each, such that it forms a closed two-dimensional figure, which is known as a circle.

A line intersecting a circle at a minimum of two distinct points is known as a secant and the line touching the circle at only one point, is known as the tangent of a circle.
The intercepted arc is the section of the circumference of a circle such that it is either encased by two chords or a line segment, meeting at a single point.

We are given that the angle subtended by secant and tangent, outside the circle is,

$\theta =90^{\circ}$

And angle measured inside the circle is,

$\phi = \theta /2\\\\\phi = 90/2\\\\\phi = 45{^\circ}$

So, the difference between the measures of the intercepted arcs is,

$\theta' = \theta - \phi\\\\\theta' = 90-45\\\\\theta' =45^{\circ}$

Thus, we can conclude that the difference between the measures of the intercepted arcs of the given circle is 45°.

learn more about the tangent here:

brainly.com/question/14022348

4 0

3 years ago

A spring is compressed by 0.0880 m and is used to launch an object horizontally with a speed of 2.76 m/s. If the object were att

lbvjy [14]

Answer:

Approximately $3.14\; {\rm rad \cdot s^{-1}}$ .

Explanation:

Fact: the angular velocity $\omega$ of a simple harmonic oscillator is the ratio between the maximum velocity $v_{\text{max}}$ and the maximum displacement $x_\text{max}$ of this oscillator. In other words:

$\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}}\end{aligned}$ .

Derivation of the previous equation:

Let $A$ denote the amplitude of this oscillation, and let $\omega$ denote the angular velocity.

The displacement of the oscillator at time $t$ would be:

$x(t) = A\, \sin(\omega\, t)$ .

The maximum displacement of this oscillator would be $x_\text{max} = A$ .

The velocity of this oscillator at time $t$ is the derivative of displacement with respect to time:

$\begin{aligned} v(t) &= \frac{d}{d t}\, [x(t)] \\ &= \frac{d}{d t} [A\, \sin(\omega\, t)] \\ &= A\, \omega\, \cos(\omega\, t)\end{aligned}$ .

The maximum velocity of this oscillator would be $v_\text{max} = A\, \omega$ .

Notice that dividing $v_\text{max} = A\, \omega$ by $x_\text{max} = A$ would give:

$\displaystyle \frac{v_\text{max}}{x_\text{max}} = \frac{A\, \omega}{A} = \omega$ .

It is given that $v_\text{max} = 2.76\; {\rm m\cdot s^{-1}}$ while $x_\text{max} = 0.0880\; {\rm m}$ . Therefore:

$\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}} \\ &= \frac{2.76\; {\rm m\cdot s^{-1}}}{0.0880\; {\rm m}} \\ &\approx 3.14\; {\rm s^{-1}}\end{aligned}$ .

(Radians per second.)

4 0

2 years ago