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Travka [436]
3 years ago
15

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

y other forces, what is the maximum acceleration a this engine can produce in 1650-kg car?
Physics
1 answer:
joja [24]3 years ago
8 0

Answer:

Approximately 9.7\; \rm m \cdot s^{-2}.

Explanation:

Assuming that there is no other force on this vehicle, the 16000\; \rm N force from the road would be the only force on this vehicle. The net force would then be equal to this 16000\; \rm N\! force. The size of the net force would be 16000\; \rm N\!\!.

Let m denote the mass of this vehicle and let \Sigma F denote the net force on this vehicle.  

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, a, would be:

\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}.

For this vehicle, \Sigma F = 16000\; \rm N whereas m = 1650\; \rm kg. The acceleration of this vehicle would be:

\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}.

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. A student times a car traveling a distance of 2 m. She finds that it takes the car 5 s to
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Show that (a)KE=1/2mv2
evablogger [386]

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{see \: below}}}}}

Explanation:

\underline{ \bold{ \sf{To \: prove \: that \: kinetic \: energy =  \frac{1}{2} m {v}^{2} }}}

Let us consider, a body of mass ' m ' is lying at rest ( initial velocity = 0 ) on a smooth surface. Let a constant force F displaces this body in its own direction by a displacement ' d '. Let 'v' be it's final velocity. The work done ' W ' by the force is given by :

\sf{W = FD}

⇒\sf{W = m \:  \times a \:  \times s} \:  \:  \:  \:  \:  \:  \:  \:  \: ( \: ∴ \: f \:  =  \: ma \: ; \: s \:  = d)

⇒\sf{W = m \:  \times  \frac{v - u}{t}  \times  \frac{u + v}{2}  \times t \:  \:  \:  \:  \:  \:  \:  \:  \: (∴ \: a =   \frac{v - u}{t} and \: s =  \frac{u + v}{2}  \times t}

⇒\sf{W = m \times  \frac{ {v}^{2}  -  {u}^{2} }{2} }

⇒\sf{W =  \frac{1}{2} m {v}^{2}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: (since, \: initial \: velocity(u) = 0)}

The work done becomes the kinetic energy of the body. Thus, the kinetic energy of a body of mass ' m : moving with the velocity equal to 'v ' is 1 / 2 mv²

∴ \sf{KE=  \frac{1}{2} m {v}^{2} }

\sf{ \underline{ \bold{  {proved}}}}

Hope I helped!

Best regards!!

5 0
3 years ago
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