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alexdok [17]
3 years ago
5

find the time taken, if the speed of a train increased from 72 km/hr to 90 km/hr for 234 km. leave your answer in seconds

Physics
1 answer:
Airida [17]3 years ago
3 0

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}

Final speed of the train, v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}

Displacement of the train, S=234\textrm{ km}=234\times 1000=234000\textrm{ m}

Using Newton's equation of motion,

v - u = at\\a=\frac{v-u}{t}

Now, using Newton's equation of motion for displacement,

v^{2}-u^{2}=2aS

Now, plug in the value of a=\frac{v-u}{t} in the above equation. This gives,

v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}

Now, plug in 234000 m for S, 25 m/s for v and 20 m/s for u. Solve for t.

t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}

Therefore, the time taken by the train is 10400 s.

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Formula used for transformer is:\frac{N_{1} }{N_{2}} = \frac{V_{1}}{V_{2}}

Where:N1 = number of turns on primary coilN2 = number of turns on secondary coilV1 = voltage on primary coilV2 = voltage on secondary coil
In a step-down transformer primary coil has more turns than secondary coil. So the ratio 1:38 means that for each turn on secondary coil we have 38 turns on primary coil.
We can solve the equation for V2:V_{2} =  \frac{ V_{1}* N_{2}  }{ N_{1} }  \\  V_{2} =  \frac{ 120* x  }{ 38x} } \\ V_{2} = 3.16V

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3 years ago
The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia. Find the moment of iner
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Answer:

See explanation

Explanation:

We have a mass m revolving around an axis with an angular speed \omega, the distance from the axis is r. We are given:

\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg]

and also the formula which states that the kinetic rotational energy of a body is:

K =\frac{1}{2}I\omega^2.

Now we use the kinetic energy formula

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After replacing in the previous equation we get:

K =\frac{1}{2}m(\omega r)^2

now we have the following:

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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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