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cupoosta [38]
3 years ago
12

When three people with a total mass of 2.00 x 102 kg step into their 1.200 x 103 kg car, the car’s

Physics
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

Answer:

a

 k =  457333.3 N/m

b

x_a  =0.09\ m      

Explanation:

From the question we are told that

    The total mass of  three people is  M  = 2.00*10^{2} \ kg

     The mass of the car is  m_c  =  1.200 *10^{3} \ kg

     The compression of the car spring is  x = 3 \ cm  = 0.03 \ m

     

Generally the spring constant is mathematically represented as

          k =  \frac{F}{x}

Here F is the force exerted by the mass of three people and that of the car , this is mathematically represented as        

=>       F = (M +m_c) *g

=>       F = ([2.0*10^{2} ]+[ 1.200*10^{3}]) * 9.8

=>       F = 13720 \  N

So

        k =  \frac{13720}{0.03}

=>     k =  457333.3 N/m

Generally if the mass which the car is loaded with is  m  =  3.00*10^{2} \ kg

Then the force experienced by the spring is  

         =>       F_a = (m +m_c) *g

         =>       F_a = (3.00*10^{3} + 1.200 *10^{3}) * 9.8

         =>       F_a = 41160 \  N

Generally from the above formula the compression is  

       x_a  = \frac{F_a}{k}

=>    x_a  = \frac{41160}{457333.3}

=>    x_a  =0.09\ m      

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Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

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