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3 years ago

What two factors affect the force of friction?

Chemistry

2 answers:

Vlad1618 [11]3 years ago

6 0

Answer : Option C) Roughness of the surface and weight of the object.

Explanation : The factors that affect the forces of friction are roughness of the surface and weight of the object.

Friction is the force which resists the relative motion of solid surfaces, fluid layers, and material elements which are sliding against each other.

Friction depends on the normal force and the roughness of the surface which is in contact. When a body is placed on the ground, the mass of the body which is weight = mass x acceleration due to gravity, comes into play for the normal force from the ground. This clearly shows that friction is affected by weight as it depends on the mass of the object.

It also depends partly on the smoothness of the contacting surfaces, if a greater force being applied to move two surfaces past one another then they are rough, if they were smooth less force would have been applied for moving it.

Sav [38]3 years ago

3 0

Roughness of the surface and weight of the object.

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29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.

arsen [322]

The answer for the following mention bellow.

Therefore the final temperature of the gas is 260 k

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 150.0 kPa

Final pressure ( $P_{2}$ ) = 210.0 kPa

Initial volume ( $V_{1}$ ) = 1.75 L

Final volume ( $V_{2}$ ) = 1.30 L

Initial temperature ( $T_{1}$ ) = -23°C = 250 k

To find:

Final temperature ( $T_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

We know;

$\frac{P*V}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ × $\frac{V_{1} }{V_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

( $P_{1}$ ) represents the initial pressure of the gas

( $P_{2}$ ) represents the final pressure of the gas

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $T_{1}$ ) represents the initial temperature of the gas

( $T_{2}$ ) represents the final temperature of the gas

So;

$\frac{150 * 1.75}{210 * 1.30}$ = $\frac{260}{T_{2} }$

( $T_{2}$ ) =260 k

Therefore the final temperature of the gas is 260 k

3 0

3 years ago

What is the binding energy of a nucleus that has a mass defect of 5.81*10-^29 kg

IrinaVladis [17]

Answer:

Choice A: Approximately $5.23 \times 10^{-29}$ joules.

Explanation:

Apply the famous mass-energy equivalence equation to find the energy that correspond to the $\rm 5.81\times 10^{-29}$ kilograms of mass.

$E = m \cdot c^{2}$ ,

where

$E$ stands for energy,
$m$ stands for mass, and
$c$ is the speed of light in vacuum.

The speed of light in vacuum is a constant. However, finding the right units for this value can simplify the calculations a lot. What should be the unit of $c$ ?

The mass given is in the appropriate SI unit:

Mass is in kilograms.

Thus, proceed with the speed of light in SI units. The SI unit for speed is meters per second. For the speed of light, $c \approx \rm 3.00\times 10^{8}\;m\cdot s^{-1}$ .

Apply the mass-energy equivalence:

$\begin{aligned} E &= m \cdot c^{2} \\ &= \rm 5.81\times 10^{-29}\; kg \times {\left(3.00\times 10^{8}\; m\cdot s^{-1}\right)}^{2}\\ &\approx \rm 5.23\times 10^{-12}\;kg\cdot m^{2}\cdot s^{-2} \end{aligned}$ .

The unit of energy is not in joules. Don't be alerted. Consider the definition of a joule of energy. One joule is the work done on an object when a force of one newton acts on the object in the direction of the force through the distance of one meter. (English Wikipedia.)

$\rm 1\; J = 1\; N \times 1\; m$ .

However, a force of one newton is defined as the force required to accelerated an object with a mass of one kilogram (not gram) at a rate of one meter per second squared. (English Wikipedia.)

$\begin{aligned}\rm 1\; J &= \rm 1\; N \times 1\; m\\ & = \rm \left(1\; kg\times 1\; m\cdot s^{-2}\right)\times 1\; m\\ &= \rm 1\; kg \cdot m^{2}\cdot s^{-2}\end{aligned}$ .

In other words, the mass defect here is also $\rm 5.23\times 10^{-12}\; J$ .

7 0

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Taya2010 [7]

Answer:

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il63 [147K]

I think the answer is B

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padilas [110]

Cell wall
Giddy Up!!!!!!!!!!!

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