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Paul [167]
2 years ago
14

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60km/h​

Physics
1 answer:
natima [27]2 years ago
4 0

The work done to stop the car is -208.33 kJ

From work-kinetic energy principles, the change in kinetic energy of the car ,ΔK equals the work done to stop the car, W.

W = ΔK = 1/2m(v'² - v²) where

  • m = mass of car = 1500 kg,
  • v = initial velocity of car = 60 km/h = 60 × 1000 m/3600 s = 16.67 m/s and
  • v' = final velocity of car = 0 m/s (since the car stops).
<h3 /><h3>Calculating the work done</h3>

Substituting the values of the variables into the equation, we have

W = 1/2m(v'² - v²)

W = 1/2 × 1500 kg(v(0 m/s)² - (16.67 m/s)²)

W = 750 kg(0 (m/s)² - 277.78 (m/s)²)

W = 750 kg(- 277.78 (m/s)²)

W = -208333.33 J

W = -208.33333 kJ

W ≅ -208.33 kJ

So, the work done to stop the car is -208.33 kJ

Learn more about work done here:

brainly.com/question/9821607

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If an electron is released at PP , what is the magnitude of the net force that these rods exert on it?
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The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.

Given values:

Length of non-conducting rod, l = 1.20 m

Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C

Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C

Distance from point P of each rod, x = 60 cm = 0.60 m

Calculation of Net electric force exerted on point P:

Consider an electron released at point P, then the net electric force exerted will be given as:

F = e. E_net       - ( 1 )

Step 1:

The net electric field value is given as:

E_net  = E₁ cos Φ + E₂ cos Φ      

           = 2E₁ cos Φ                  -( 2 )

where, E₁ & E₂ are electric fields due to positive and negative rod                

            respectively.

            Φ is phase angle

Step 2:

The electric field due to positive rod is given as:

E₁ = k (λ/r)             - ( 3 )

where, k is Coulomb's force constant

            λ is linear charge density

            r is distance between point P and half of the rod.

Now, the linear charge density is given as:

λ = Charge/length = Q/x

The value of r is given as:

r = √x²-a²

where, x is length of rod

           a is half length of rod

Applying values in above equation, we get:

r = √x²-(x/2)²

r = √(1.20 m)²-(1.20/2)²

  = √1.08

  = 1.04 m

Substituting all the determined values in equation 3 we get:

E₁ = k (λ/r)

   = k [(Q/x)/r]

   = k [ Q/xr ]

   = (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

   = 1.803×10⁴ N/C

Step 3:

Similarly, the electric field due to negative rod is given as:    

E₂ = k [ Q/xr ]

    = (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

    = 1.803×10⁴ N/C

Step 4:

Consider equation 2:

E_net  = 2E₁ cos Φ

From the figure we get the phase angle as:

Φ = tan⁻¹ (0.60 m/0.60 m)

   = tan⁻¹ ( 1 )

   = π/4  

Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:

E_net = 2(1.803×10⁴ N/C) cos π/4

          = 2(1.803×10⁴ N/C) (0.5)

          = 18030 N/C

Step 5:

Consider equation 1 :

F = e. E_net

where, e is charge on an electron

Applying values in above equation we get:

F = (1.6 × 10⁻¹⁹ C)(18030 N/C)

  = 2.885 × 10⁻¹⁵ N

Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is  2.885 × 10⁻¹⁵ N.

Learn more about electric force here:

<u>brainly.com/question/1634182</u>

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