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3 years ago

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B. A 20kg wagon is pulled up a 30° incline at an acceleration of 2.5ms?. The force pulling the wagon is parallel to the incline

and its value is 182N. Determine the value of the co-efficient of kinetic friction. [5]

1 answer:

yawa3891 [41]3 years ago

7 0

The value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.

<h3>Coefficient of the kinetic friction</h3>

The value of coefficient of kinetic friction is calculated as follows;

F - Ff = ma

F - μmgcosθ = ma

where;

F is applied force
μ is coefficient of kinetic friction
m is mass of the wagon
a is acceleration of the wagon

182 - μ(20 x 9.8 x cos30) = 20(2.5)

182 - 169.74μ = 50

182 - 50 = 169.74μ

132 = 169.74μ

μ = 132/169.74

μ = 0.78

Thus, the value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.

Learn more about coefficient of friction here: brainly.com/question/20241845

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3 years ago

How do you answer this. Need help ASAP. Offering 20 points !!!

Temka [501]

in this since your volume remains at a constant you'll need to use Gay-Lussacs law, p1/t1=p2/t2.

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just add 273 to your celcius

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then you set up the equation

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distribute the 315 to the pressure.

9.45×10^8=270x then you divide 270 o both sides to get

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3 years ago

Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.

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An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

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Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

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3 years ago

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In our problem, the charge is q=2 C, and the force experienced by it is
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3 years ago