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3 years ago

13

What are the factors that effect the strength of a frictional force???

1 answer:

seropon [69]3 years ago

5 0

If you write down the formula for friction, you will get an answer.

Ff = u * N Where N is a push down force that an object experiences.

u (mu) is a constant and has no units

It may not be accelerating and still experience friction. A is not correct.

Color and Density will not affect the frictional force. B is not so.

Buoyant forces are a different thing altogether. Generally friction has nothing to do with them. C is incorrect.

The last one is your answer. Technically mg should be the answer and not mass, but the second part is correct.

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At positions and_______, _______kinetic energy is maximum. At position

worty [1.4K]

Hi there!

I believe it looks like this.

At positions A, G kinetic energy is maximum.
At position D, potential energy is maximum.

Hope this helps! ☺♥

7 0

3 years ago

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An object is moving with uniform speed in a circle of radius r. Calculate the distance and displacement

dimaraw [331]

Answer and Explanation:

distance will be 2×3.14 (pie)×r

displacement will be 2r (diameter)

the motion is uniform circular motion as the object is moving in a circular path with uniform motion

8 0

2 years ago

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The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t

EastWind [94]

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

$Ay = By = \frac{w * 6}{2} = 3w$

$P_c_r = 3w * F.S = 3w * 2.0 = 6w$

$I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7$

To find the maximum intensity, w, let's take the Pcr formula, we have:

$P_c_r = \frac{\pi^2 E I}{(KL)^2}$

Let's take k = 1

$E = 200*10^9$

Substituting figures, we have:

$6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}$

Solving for w, we have:

$w = \frac{67266.84}{6}$

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

$\sigma _c_r = \frac{w}{A}$

$\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y$ . This means it is safe

The maximum intensity w = 11.211KN/m

3 0

3 years ago

The table below shows the approximate distances of two stars from Earth.

Gemiola [76]

I think that the answer is c. good luck

6 0

3 years ago

Find the magnitude of the sum

shusha [124]

Answer:

$\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m$

Explanation:

<u>Sum of Vectors in the Plane</u>

Given two vectors

$\displaystyle \vec{v_1}\ ,\ \vec{v_2}$

They can be expressed in their rectangular components as

$\displaystyle \vec{v_1}=$

$\displaystyle \vec{v_2}=$

The sum of both vectors can be done by adding individually its components

$\displaystyle \vec{v_1}+\vec{v_2}=$

If the vectors are given as a magnitude and an angle $(M\ ,\ \theta )$ , each component can be found as

$\displaystyle \vec{v_1}=$

$\displaystyle \vec{v_2}=$

The first vector has a magnitude of 3.14 m and an angle of 30°, so

$\displaystyle \vec{v_1}=$

$\displaystyle \vec{v_1}=$

The second vector has a magnitude of 2.71 m and an angle of -60°, so

$\displaystyle \vec{v_2}=$

$\displaystyle \vec{v_2}=$

The sum of the vectors is

$\displaystyle \vec{v_1}+\vec{v_2}=$

$\displaystyle \vec{v_1}-\vec{v_2}=$

Finally, we compute the magnitude of the sum

$\displaystyle |\vec{v_1}+\vec{v_2}|=\sqrt{(4.08)^2+(-0.78)^2}$

$\displaystyle |\vec{v_1}+\vec{v_2}|=\sqrt{17.25}$

$\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m$

3 0

3 years ago