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Vikentia [17]
2 years ago
5

The CM of an empty 1300-kg car is 2.45 m behind the front of the car.

Physics
1 answer:
nydimaria [60]2 years ago
7 0

The center of mass of the car with all people seated will be 3.20 m far from the front of the car.

Taking the front of the car as reference:

m₁x₁ + m₂x₂ + m₃x₃ = mx

where,

m₁ = mass of empty car = 1300 kg

x₁ = distance of center of mass of empty car from front = 2.45 m

m₂ = mass of 2 people sitting in front = 2 x 60 kg = 120 kg

x₂ = distance of center of mass of 2 people sitting in front from the front    = 2.7 m

m₃ = mass of 3 people sitting in back seat = 3 x 60 kg = 180 kg

x₃ = distance of center of mass of 3 people sitting in the back seat from the front = 3.65 m

m = total mass of car with all people = 1300 kg + 120 kg + 180 kg = 1600kg

x = distance of center of mass of the car from the front when all are seated

Therefore,

(1300 kg)(2.45 m) + (120 kg)(2.7 m) + (180kg)(3.65 m) = (1300 kg)x

x = 4166 kg.m/1300 kg

x = 3.20 m

Therefore, the center of mass of the car with all people seated will be 3.20 m far from the front of the car.

Learn more about center of mass here:

brainly.com/question/17088562

#SPJ1

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lukranit [14]

Answer:

Mass of the steel cube =  7800 kg

Volume of the steel  = 1.025 cubic centimetre

Explanation:

Given:

The density of the steel   =  7.8

Side of the cube = 12 cm

<u>(1)The mass of steel cube :</u>

We know that,

Density  = \frac{Mass}{Volume}

We are given with density and sides of the cube

then volume of the cube

= (side)^3

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Now

7.8 = \frac{mass}{1000}

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<u>(2)volume of steel:</u>

Given the mass  = 8 kg

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Substituting the values

7.8 = \frac{8}{volume}

volume = \frac{8}{7.8}

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\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}
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                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

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