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3 years ago

Which material is used to reduce friction ?

Physics

1 answer:

Iteru [2.4K]3 years ago

6 0

Answer:MOST COMMON METHOD IS USING A LUBRICANT -

A lubricant is a substance, usually organic, introduced to reduce friction between surfaces in ... medical examination. It is mainly used to reduce friction and to contribute to a better and efficient functioning of a mechanism. ... For lubricant base oil use, the vegetable derived materials are preferred

OTHER METHODS-

There are a number of ways to reduce friction:

Make the surfaces smoother. ...

Lubrication is another way to make a surface smoother.

Make the object more streamlined.

Reduce the forces acting on the surfaces.

Reduce the contact between the surfaces.

PLEASE DO MARK ME AS THE BRAINLIEST :)

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A rough surface for infrared waves may be polished for

zloy xaker [14]

Answer:

Radio Waves

Explanation:

7 0

2 years ago

A metal container with the oil weigh

MA_775_DIABLO [31]

Answer:

16 kg

Explanation:

M - container

m - oil mass

by definition of density ,

relative density is the ratio of the density of a substance to the density of water.

relative density = density/ density of water

density of oil = 1.2*1000 kgm⁻³ = 1200 kgm⁻³

1 Litre =10⁻³ m³

oil volume = 80 *10⁻³ m³

mass of oil = density * volume

= 1200*80*10⁻³

= 96 kg

Mass of container + mass of oil =112

mass of container = 112 - 96

= 16 kg

7 0

3 years ago

A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl

Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

4 0

3 years ago

Which of the following is an example of deposition?

Murrr4er [49]

B would be the correct answer

4 0

3 years ago

A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end

Natalka [10]

Answer:

The angular acceleration α = 14.7 rad/s²

Explanation:

The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod

Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

So Iα = Wr

Substituting the value of the variables, we have

mL²α/12 = mgL/2

Simplifying by dividing through by mL, we have

mL²α/12mL = mgL/2mL

Lα/12 = g/2

multiplying both sides by 12, we have

Lα/12 × 12 = g/2 × 12

αL = 6g

α = 6g/L

α = 6 × 9.8 m/s² ÷ 4.00 m

α = 58.8 m/s² ÷ 4.00 m

α = 14.7 rad/s²

So, the angular acceleration α = 14.7 rad/s²

5 0

3 years ago