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A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 106.0 km/h. At that speed, the forces resisting motion, including friction and air resistance, total 432.0 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force that motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 241 kg

Sagot :

Cricetus

Answer:

"1155 N" is the appropriate solution.

Explanation:

Given:

Acceleration,

[tex]a=3 \ m/s^2[/tex]

Forces resisting motion,

[tex]F_f=432 \ N[/tex]

Mass,

[tex]m = 241 \ kg[/tex]

By using Newton's second law, we get

⇒ [tex]F-F_f=ma[/tex]

Or,

⇒         [tex]F=ma+F_f[/tex]

By putting the values, we get

⇒             [tex]=(3\times 241)+432[/tex]

⇒             [tex]=723+432[/tex]

⇒             [tex]=1155 \ N[/tex]

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