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3 years ago

Write some harms of friction.

Physics

2 answers:

Serjik [45]3 years ago

7 0

Answer:

Friction had many disadvantages and these are the har,s of friction too.

Explanation:

Disadvantages of friction are mainly forest fires, decreasing life expectancy of vehicles, damages many machines, generates heat, wear and tear, wastage of force/energy, etc. When two things in a forest rub against each other through an amount of force, friction will be present and it will cause a spark. This spark will be continued into a fire because of other materials like wood in the forest causing a forest fire. Secondly, vehicles work through moving many internal machines in them; this again includes friction and friction can damage the machines so to prevent it ppl grease the machines. decreasing life of vehicles is just a larger view of damaging the machines as this disadvantage of friction can be used anywhere. Wastage of force/ energy because when we roll a ball on a surface we have to put force on the ball to make it move forward but because friction moves in the opposite direction energy/force will be wasted there. Generating heat and wear & tear are some factors of friction which is also a harm.

Veseljchak [2.6K]3 years ago

3 0

Answer:

Friction produces unnecessary heat leading to the wastage of energy. The force of friction acts in the opposite direction of motion, so friction slows down the motion of moving objects. Forest fires are caused due to the friction between tree branches. However, friction can also cause problems in a car. Friction between moving engine parts increases their temperature and causes the parts to wear down. Friction can be both harmful and helpful, so it may be necessary to decrease or increase friction.

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6. An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of t

zubka84 [21]

Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

$d=v\times t$

$d=9.1\times t$ ....(I)

The distance of longitudinal wave

$d=5.7\times (t+71)$ ....(II)

From the first equation

$t=\dfrac{d}{9.1}$

Put the value of t in equation (II)

$d =5.7\times(\dfrac{d}{9.1}+71)$

$\dfrac{9.1d-5.7d}{9.1}=71\times5.7$

$d0.3736=404.7$

$d =1083.24\ km$

Hence, The distance away the center of the earthquake is 1083.24 km.

6 0

3 years ago

Suppose the electrons and protons in 1g of hydrogen could be separated and placed on the earth and the moon, respectively. Compa

MAXImum [283]

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

$F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}$

Where:

G= 6.674×10^−11 N · (m/kg)2

k = 8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

$\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}$ --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is 5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

$\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}$

Therefore

$\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}$

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

7 0

3 years ago

The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as , where a

Misha Larkins [42]

Answer:

3 times louder

Explanation:

The Loudness in decibel Db L = 10㏒(I/I₀) where I = sound intensity level and I₀ = threshold of hearing = 10⁻¹² W/m².

Now, for Jessica, I₁ = sound intensity level of Jessica's music = 10⁻⁹

and I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₁ = 10㏒(I₁/I₀)

L₁ = 10㏒(10⁻⁹/10⁻¹²)

L₁ = 10㏒(10³)

L₁ = 3 × 10㏒10

L₁ = 30㏒10

L₁ = 30 dB

Now, for Braylee, I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₂ = 10㏒(I₁/I₀)

L₂ = 10㏒(10⁻³/10⁻¹²)

L₂ = 10㏒(10⁹)

L₂ = 9 × 10㏒10

L₂ =90㏒10

L₂ = 90 dB

So, the number of times Braylee's music is louder than Jessica's music is L₂/L₁ = 90 dB/30 dB = 3

So, Braylee's music is 3 times louder than Jessica's music

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3 years ago

A bus travels east for 3 km, then north for 4 km. What is its final displacement?

ikadub [295]

5 km northeast. Left and up would make northeast

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3 years ago

The distance from the earth to a star that has identical apparent and absolute magnitudes is ____ .

PSYCHO15rus [73]

I think it's distance.

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