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2 years ago

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Which of the following is not a factor in whether a reaction will spontaneously occur? A. Entropy change of the system B. Enthal

py change of the system C. Temperature of the system D. Pressure change of the system

1 answer:

Delvig [45]2 years ago

6 0

Answer:

D

Explanation:

pressure change have nothing to do with the spontaneity.

Entropy change , enthalpy change , temperature have roles in deciding spontaneity.

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As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road i

Kitty [74]

Answer:

<em>B</em><em>.</em><em> </em><em>Kinetic</em><em> </em><em>friction</em><em> </em>

Explanation:

This is definitely the correct answer because kinetic friction acts when an object is in motion and it allows the object to move without slipping, etc

<em>ALSO</em><em>,</em><em> </em><em>PLEASE DO</em><em> </em><em>MARK</em><em> </em><em>ME AS</em><em> </em><em>BRAINLIEST UWU</em><em> </em>

<em>Bonne</em><em> </em><em>journée</em><em> </em><em>;</em><em>)</em><em> </em>

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2 years ago

Read 2 more answers

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : <em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

part b : <em>The void ratio is 0.77</em>

part c : <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

<em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

2 years ago

A snowball is melting at a rate of 324π mm3/s. At what rate is the radius decreasing when the volume of the snowball is 972π mm3

Oduvanchick [21]

Answer:

The radius is decreasing at 4 mm/s

Explanation:

The volume of a sphere is:

$V = 4/3*\pi *r^3$ So, when the volume is 972π mm^3 the radius r is:

r = 9mm

Now, the change rate is given by the derivative:

$dV/dt = 4/3*\pi *3*r^2*dr/dt$

Where: dV/dt = -324π mm^2/s

r = 9mm

Solving for dr/dt:

dr/dt = -4mm/s

5 0

2 years ago

Why does carpet tend to produce differences in static electricity more that hardwood or tile floors

Makovka662 [10]

Answer:

This is because the rubbing releases negative charges, called electrons, which can build up on one object to produce a static charge. For example, when you shuffle your feet across a carpet, electrons can transfer onto you, building up a static charge on your skin.

Explanation:

This is because the rubbing releases negative charges

4 0

2 years ago

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

yanalaym [24]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

That would be our first equation

Now, we need to analyze its vertical motion

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Knowing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Solving for t

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

So, the magnitude of the velocity is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The <u><em>direction of the ball</em></u> is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

4 0

2 years ago