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gulaghasi [49]
2 years ago
6

A horizontal force, F, acts on a crate weighing 8 N, which moves on a horizontal floor at constant speed. The coefficient of fri

ction between the crate and the surface is 0.25. The work done by F when the crate has moved 12 m is: Group of answer choices 2 J None of the above 24 J 96 J 0 J 10 J
Physics
1 answer:
RoseWind [281]2 years ago
7 0

The answer is 24 J

F K =.25*8 N

     = 2N

F = f k = 2 N

Since a = 0

W = f * s

 2 N * 12 m = 24 J

The coefficient of friction is a ratio used to quantify the friction force among two gadgets when it comes to the everyday pressure this is keeping them collectively. The coefficient of friction is critical attention at some stage in material selection and floor requirement determination.

For instance, ice on steel has a low coefficient of friction – the 2 materials slide past each different without problems – whilst rubber on the pavement has an excessive coefficient of friction – the substances no longer slide past each other without difficulty.

The coefficient of friction is dimensionless and it does not have any unit. it is a scalar, meaning the direction of the force does not have an effect on the physical quantity. The coefficient of friction depends on the gadgets that are causing friction.

Learn more about the coefficient of friction here brainly.com/question/20241845

#SPJ4

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Answer:

A metalloid is used because it is a semiconductor and can become more conductive when more light shines on it

Explanation:

The material used in a solar panel is a metalloid. It can often become conductive when more light shines on it.

Metalloids have properties that straddles between those of metals and non-metals.

In essence, they can be conductive or not under certain conditions.

The most important property they exhibit is that they can become more conductive when more light shines on them. This way more electrons are produced.

3 0
3 years ago
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A yo-yo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be taken to be MR2/2 and the thickn
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Answer:

The tension in the cord is T=\frac{MR^{2}g }{2b^{2}+R^{2}  }

Explanation:

Given:

M = mass

b = radius

R = spool of radius

The equation is:

bT=(\frac{MR^{2} }{2} )(\frac{a}{b} )\\T=\frac{MR^{2}a }{2b^{2} } (eq. 1)

The sum of forces in y:

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Mg=(M+\frac{MR^{2} }{2b^{2} }  )a\\a=\frac{2b^{2}g }{2b^{2}+R^{2}  }

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T=\frac{MR^{2} }{2b^{2} } (\frac{2b^{2}g }{2b^{2} +R^{2} } )\\T=\frac{MR^{2}g }{2b^{2}+R^{2}  }

3 0
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Determine the slope of end a of the cantilevered beam. E = 200 gpa and i = 65. 0(106) mm4
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For E = 200 gpa and i = 65. 0(106) mm4,  the slope of end a of the cantilevered beam  is mathematically given as

A=0.0048rads

<h3>What is the slope of end a of the cantilevered beam?</h3>

Generally, the equation for the   is mathematically given as

A=\frac{PL^2}{2EI}+\frac{ML}{EI}

Therefore

A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}

A=0.00288+0.00192=0.0048rads

A=0.0048rads

In conclusion,  the slope is

A=0.0048rads

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Answer:

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