All categories

3 years ago

How do you think the mass of an object affects how much friction get exerted on the object by a rough surface?

Physics

1 answer:

Shkiper50 [21]3 years ago

7 0

Explanation:

Friction can't affect mass of a substance(considering for a substance whose mass doesn't change with time),rather it is the mass of an object which can affect friction variously. Let's take some example to understand the situation. So,higher the mass of the object,higher will be the frictional force.

Friction occurs because no surface is perfectly smooth. Rougher surfaces have more friction between them. Heavier objects also have more friction because they press together with greater force. Friction produces heat because it causes the molecules on rubbing surfaces to move faster and have more energy.

You might be interested in

If the kinetic energy of an electron is 4.1e-18 j, what is the speed of the electron? (you can use the approximate (nonrelativis

arlik [135]

The kinetic energy of the electron is
$K= \frac{1}{2}mv^2$
where $m=9.1 \cdot 10^{-31} kg$ is the mass of the electron and v its speed. Since we know the value of the kinetic energy, $K=4.1 \cdot 10^{-18} J$ , we can find the value of the speed v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 4.1 \cdot 10^{-18}J}{9.1 \cdot 10^{-31}kg} } = 3\cdot 10^6 m/s$

3 0

4 years ago

Percent Yield Lab Report

Vlada [557]

Based on the data obtained from the reaction, the following conclusions can be made;

according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
the percent yield is less than 100% because of loss in mass of either reactants or products.
the errors could have occurred during the weighing and transfer of reactants and products
repeated measurements are required in order to improve accuracy

<h3>What is the percent yield of the reaction?</h3>

Equation of the reaction is given below:

2 Mg + O₂ ----> 2 MgO

Trial 1

Mass of empty crucible with lid = 26.698 (g)

Mass of Mg metal, crucible, and lid = 27.040 (g)

Mass of MgO, crucible, and lid = 27.198 (g)

Mass of metal = 27.040 - 26.698 = 0.342

Mass of MgO = 27.198 - 26.698 = 0.500

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

molar mass of Mg = 24 g

moles of Mg = 0.342/24 = 0.01425

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.01425

moles of MgO produced = mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.500/40 = 0.0125

<h3>Percent yield</h3>

Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percentage yield = 0.0125/0.01425 * 100%

Percent yield of MgO = 87.7%

Trial 2

Mass of empty crucible with lid = 26.691 (g)

Mass of Mg metal, crucible, and lid = 27.099 (g)

Mass of MgO, crucible, and lid = 27.361 (g)

Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

molar mass of Mg = 24 g

moles of Mg = 0.408/24 = 0.0170

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.0170

moles of MgO produced = mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.670/40 = 0.01675

<h3>Percent yield</h3>

Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percent yield = 0.01675/0.0170 * 100%

Percent yield of MgO = 98.5%

Average percent yield = (87.7 + 98.5)% / 2

Average percent yield = 89.0%

Based on the data obtained from the reaction, the following conclusion can be made;

according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
the percent yield is less than 100% because of loss in mass of either reactants or products.
the errors could have occurred during the weighing and transfer of reactants and products
repeated measurements are required in order to improve accuracy

Learn more about law of conservation of mass at: brainly.com/question/1824546

6 0

3 years ago

A man standing 1.54 m in front of a shaving mirror produces a real, inverted image 15.2 cm from it. What is the focal length of

zavuch27 [327]

Answer:

The focal length is 16.86 cm and the distance of the man if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

Explanation:

Given that,

Object distance u=1.54 m =154 cm

Image distance v = 15.2 cm

Magnification = 2

We need to calculate the focal length

Using formula of mirror

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{15.2}+\dfrac{1}{-154}$

$\dfrac{1}{f}=\dfrac{347}{5852}$

$f=16.86\ cm$

We need to calculate the focal length

Using formula of magnification

$m= \dfrac{-v}{u}$

Put the value into the formula

$2=\dfrac{v}{u}$

$v = -2u$

Using formula of for focal length

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{16.86}=\dfrac{1}{u}-\dfrac{1}{2u}$

$\dfrac{1}{16.86}=\dfrac{1}{2u}$

$2u=16.86$

$u=\dfrac{16.86}{2}$

$u=8.43\ cm$

Hence, The focal length is 16.86 cm and the distance of the man if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

3 0

4 years ago

Which term means that humans are combining two different organism's DNA to create a new organism. Hint: Oink, Oink Question 3 op

cestrela7 [59]

Answer:

breeding

Explanation:

3 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago