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3 years ago

Friction is generated when _____ interact with each other on sliding surfaces.

Physics

2 answers:

Hatshy [7]3 years ago

8 0

Answer:

Friction is generated when molecules interact with each other on sliding surfaces.

Explanation:

As we know that friction force is the contact force between two surfaces. When two surface are placed over each other then in that case molecules of one surface are in contact with the molecules of other surface.

Due to the close interaction all the molecules will form a weak attraction force which will make them bonded with each other.

Now due to this weak bond when we move one surface relative to other surface then we need the force to break the bond between two surface which is known as friction force.

So here we can say that

Friction is generated when molecules interact with each other on sliding surfaces.

vladimir1956 [14]3 years ago

7 0

Friction is generated when 2 solid surfaces interact

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Two frisky grasshoppers collide in midair at the top of their respective trajectories and grab onto each other, holding tight th

Svetlanka [38]

Answer:

The decrease in Kinetic energy is 0.0107 Joules

Explanation:

Given

Mass of grasshoppers

Let m1 = Mass of grasshopper 1

Let m2 = Mass of grasshopper 2

Let u1 = initial speed of grasshopper 1

Let u2 = initial speed of grasshopper 2

m1 = 250g = 0.25kg

m2 = 130g = 0.13kg

u1 = 15cm/s = 0.15m/s

u2 = 65cm/s = 0.65m/s

First, we calculate the final velocity of the grasshoppers after collision using conservation of momentum.

Using

m1u1 + m2u2 = (m1 + m2) * v

Where v = final velocity

By substituton

0.25 * 0.15 + 0.13 * 0.65 = (0.25 + 0.13) * v

0.0375 + 0.0845 = 380v

0.122 = 0.38v

Make v the subject of formula

v = 0.122/0.38

v = 0.321 m/s

Calculating the Kinetic energies before and after impact.

Before collision;

KE = ½m1u1²+ ½m2u2²

KE = ½(m1u1² + m2u2²)

By substituton;

KE = ½(0.25 * 0.15² + 0.13 * 0.65²)

KE = 0.030275J

After collision:

KE = ½(m1+m2)v²

KE = ½(0.25 + 0.13) * 0.321²

KE = 0.01957779 J

Change in kinetic energy = ∆KE

∆KE = 0.030275J - 0.01957779J

∆KE = 0.01069721J

∆KE = 0.0107 J --- Approximately

Hence the decrease in Kinetic energy is 0.0107 Joules

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3 years ago

For which type(s) of signals does noise alter the original information?

nika2105 [10]

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3 years ago

a person can throw a 300 gram ball at 25 m/s the maximum height the person can throw the ball on Earth?

jolli1 [7]

Answer:

Elevation =31.85[m]

Explanation:

We can solve this problem by using the principle of energy conservation. This consists of transforming kinetic energy into potential energy or vice versa. For this specific case is the transformation of kinetic energy to potential energy.

We need to first identify all the input data, and establish a condition or a point where the potential energy is zero.

The point where the ball is thrown shall be taken as a reference point of potential energy.

$E_{p} = E_{k} \\where:\\E_{p}= potential energy [J]\\ E_{k}= kinetic energy [J]$

m = mass of the ball = 300 [gr] = 0.3 [kg]

v = initial velocity = 25 [m/s]

$E_{k}=\frac{1}{2} * m* v^{2} \\E_{k}= \frac{1}{2} * 0.3* (25)^{2} \\E_{k}= 93.75 [J]$

$93.75=m*g*h\\where:\\g = gravity = 9.81 [m/s^2]\\h = elevation [m]\\replacing\\h=\frac{E_{k}}{m*g} \\h=\frac{93.75}{.3*9.81} \\h=31.85[m]$

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4 years ago

The products of a chemical reaction can never have greater mass than the reactants

jek_recluse [69]

The answer to this is true

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4 years ago

Read 2 more answers

2. Who bombarded molybdenum with atomic number (Z=42) with fast moving neutrons?

Dominik [7]

Answer:

This question appears incomplete

Explanation:

This question appears incomplete, however Molybdenum-99 (⁹⁹Mo) is produced by bombarding Molybdenum-98 (⁹⁸Mo) with fast moving neutrons (¹₀n) as shown below

⁹⁸₄₂Mo + ¹₀n ⇒ ⁹⁹₄₂Mo + ⁰₀γ

This reaction is a nuclear caption reaction (which occurs in a nuclear reactor) for the production of Molybdenum-99 (⁹⁹Mo) which serves as a "precursor" for the production of medically viable/ Clinical Grade Technutium-99m (⁹⁹Tc) through Ion-exchange technique.

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4 years ago