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alexdok [17]
3 years ago
11

during a crash, the acceleration was less than 30g. Calculate the force on a 70 kg person accelerating at this rate.

Physics
1 answer:
Igoryamba3 years ago
4 0

Answer:

less than 20580 N

Explanation:

According to the newton's second law of motion

Force = mass * acceleration

(assuming gravitational acceleration =9.8 m/s2 )

acceleration = 30*9.8 = 294 m/s2

acting Force = 70 * 294

                     = 20580 N

Since the acceleration was less than 30g  , acting force should also be less than 20580 N

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Answer:

The total resistance of the wire is = 1.917\times10^{-3}

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, R_{Total}

\frac{1}{R_{Total}}  =\frac{1}{R_{cu}  }+\frac{1}{R_{al}}

Parameters given:

Length of wire = 1 m

Cross sectional area of copper A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3}  )^{2} =3.142\times10^{-6} m^{2}

Cross sectional area of aluminium wire  

A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3}  )^{2}-(1\times 10^{-3}  )^{2}] =9.42\times10^{-6} m^{2}\\

Resistivity of copper \rho _{cu}=1.7\times 10^{-8}  \Omega .m

Resistivity of Aluminium \rho _{al}=2.8\times 10^{-8}  \Omega .m

Resistance of copper R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} }  =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega

Resistance of aluminium R_{al}= \frac{\rho_{al} \times l}{A_{al} }  =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega

The total resistance of the wire can be obtained as follows;

\frac{1}{R_{Total}}  =\frac{1}{5.41\times10^{-3}  }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}

R_{Total}= 1.917\times 10^{-3}\Omega

∴ The total resistance of the wire = 1.917\times 10^{-3}\Omega

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A free-falling object experiences a downward acceleration of 9.8 m/s/s (on Earth). This specific designation is given to the numerical value for an object in free fall because it is such an essential value. The longer an object is in free fall, the faster it descends toward the ground due to gravity. In actuality, an object's velocity rises by 9.8 m/s2, so it reaches 9.8 m/s by the time it begins to fall.

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Answer:

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(b) The image is virtual and magnified.

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