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The sensor in the torso of a crash test dummy records the magnitude and direction of the net force acting on the dummy.If the dummy is thrown forward with a force of 130.0N while simultandously being hit from the side wigh a force of 4500.0N,what force will the sensor report

Respuesta :

Explanation:

Given that,

The dummy is thrown forward with a force of 130 N, [tex]F_1=130\ N[/tex]

Side force acting on the dummy, [tex]F_2=4500\ N[/tex]

We need to find the force acting on the sensor report. It can be calculated using Pythagoras theorem as :

[tex]F_{net}=\sqrt{F_1^2+F_2^2}[/tex]

[tex]F_{net}=\sqrt{130^2+4500^2}[/tex]

[tex]F_{net}=4501.87\ N[/tex]

So, the net force acting on the sensor report is 4501.87 N. Hence, this is the required solution.

The net force acting on the dummy is 4502 N.

Force is a vector, the resultant force (net force) is that single force that has the same effect in magnitude and direction as two or more forces acting together. The resultant of a vector must take into cognizance, the geometry of the problem.

The dummy is thrown forward with a force of 130.0N while simultaneously being hit from the side with a force of 4500.0N. The net force must now be obtained by Pythagoras theorem.

Fnet^2 = F1^2 + F2^2

F1 = 130.0N

F2 = 4500.0N

Fnet = √(130.0N)^2 + (4500.0N)^2

Fnet = 4502 N

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