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3 years ago

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Describe all the ways that newtons laws can apply in a car crash

1 answer:

Harman [31]3 years ago

7 0

Newtons first law - Objects in the car at rest (The human) will remain at rest unless affected by an unbalanced force. Well the unbalanced force would be the crash and this would set the human in motion and they would ether fly out the car if not wearing a seat belt or if wearing one they would get bad whip lash

Newtons second law - With more mass requires more force, so since the human is pretty light or even if heavy in a big crash there will be so much more from it that this will send the human flying.

Newtons 3rd law - Objects A puts force onto objects b and object b excretes the same amount of force back onto object a, so in a crash the human would hit the car hard and the car would excrete the same amount of force back on the human which would really damage him/her

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Option (2)

Explanation:

From the figure attached,

Horizontal component, $A_x=A\text{Sin}37$

$A_x=12[\text{Sin}(37)]$

= 7.22 m

Vertical component, $A_y=A[\text{Cos}(37)]$

= 9.58 m

Similarly, Horizontal component of vector C,

$C_x$ = C[Cos(60)]

= 6[Cos(60)]

= $\frac{6}{2}$

= 3 m

$C_y=6[\text{Sin}(60)]$

= 5.20 m

Resultant Horizontal component of the vectors A + C,

$R_x=7.22-3=4.22$ m

$R_y=9.58-5.20$ = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

$R=\sqrt{(R_x)^{2}+(R_y)^2}$

= $\sqrt{(4.22)^2+(4.38)^2}$

= $\sqrt{17.81+19.18}$

= 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = $\frac{\text{CB}}{\text{OB}}$

= $\frac{R_y}{R_x}$

= $\frac{4.38}{4.22}$

m∠COB = $\text{tan}^{-1}(1.04)$

= 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.

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