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Dennis_Churaev [7]
3 years ago
15

A car starts from rest and accelerates uniformly to a speed of 72 km per hour over a distance of 500m. calculate acceleration​

Physics
1 answer:
Kisachek [45]3 years ago
6 0

Answer:

0.4 m/s^2 or 5184 km/h^2

Explanation:

To find the acceleration, given the inicial and final speed of the car, we can use the Torricelli's formula:

V^2 = Vo^2 + 2*a*D

Where V is the final speed, Vo is the inicial speed, a is the acceleration and D is the distance travelled. So, with V = 72 km/h, Vo = 0 (starts from rest) and D = 0.5 km, we have:

72^2 = 0 ^2 + 2*a*0.5

a = 72^2 = 5184 km/h^2

For an acceleration in m/s^2, we should use the speed in meters per second (72 / 3.6 = 20 m/s) and the distance in meters (D = 500):

20^2 = 0^2 + 2*a*500

400 = 1000a

a = 0.4 m/s^2

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Answer:

k = 9.6 x 10^5 N/m or 9.6 kN/m

Explanation:

First, we need to use the expression to calculate the spring constant which is:

w² = k/m

Solving for k:

k = w²*m

To get the angular velocity:

w = 2πf

The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:

f = V/x

f = 5.7 / 4.9 = 1.16 Hz

Now the angular velocity:

w = 2π*1.16

w = 7.29 rad/s

Finally, solving for k:

k = (7.29)² * 1800

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3 years ago
What is a literature review?<br>​
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Given one mole of diamond vs one mole of graphite,
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Answer:

The pressure is P= -  6.39*10^8Pa

The temperature is T =1218.63 K

Explanation:

Generally Gibbs free energy is mathematically represented as

                   G = E + PV -TS

   Where  E is the enthalpy

               PV is the pressure volume energy (i.e PV energy)

                S  is the entropy

                T is the temperature

For stability to occur the Gibbs free energy must be equal to zero

Considering Diamond

  So at temperature of  T = 300 K

         E + PV - TS = 0

making P the subject

          P = \frac{TS-E}{V}

Now substituting 300 K for T , 2900 J  for E ,

                              3.42cm^3 = \frac{3.42}{1*10^6} = 3.42*10^{-6}m^3 for V and 2.38 J/K for S

     P = \frac{(300 * 2.38)- 2900}{3.42*10^{-6}}

         P= -  6.39*10^8Pa

The negative sign signifies the direction of the pressure

Given that  P = 1*0^5Pa

making T the subject

            T = \frac{PV+E}{S}

Substituting into the equation

            T = \frac{1*10^5 * 3.42 *10^{-6}+2900}{2.38}

                T =1218.63 K

             

         

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