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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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3 years ago
A car traveling at 30 m/s drives off a cliff that is 50 meters high? How far away does it land?
Semenov [28]

Answer:

The maximum range R_{max}= 132. 72 m

Explanation:

Given,

The initial velocity of the car, u = 30 m/s

The height of the cliff, h = 50 m

Let the car drives off the cliff with a horizontal velocity of 30 m/s.

The formula for a projectile that is projected from a height h from the ground is given by the relation

                                R_{max}= \frac{u}{g}\sqrt{u^{2} + 2gh }  m

Where,

                          g - acceleration due to gravity

Substituting the values in the above equation

                   R_{max}= \frac{30}{9.8}\sqrt{30^{2} + 2X9.8X50 }  

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3 years ago
A charge -353e is uniformly distributed along a circular arc of radius 5.30 cm, which subtends an angle of 48°. What is the line
vladimir2022 [97]

Answer:

- 1.3 x 10⁻¹⁵ C/m

Explanation:

Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C

r = Radius of the arc = 5.30 cm = 0.053 m

θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad        (Since 1 deg = 0.0175 rad)

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length of the arc is given as

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L = (0.053) (0.84)

L = 0.045 m

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Linear charge density is given as

\lambda =\frac{Q}{L}

Inserting the values

\lambda =\frac{-564.8\times 10^{-19}}{0.045}

λ = - 1.3 x 10⁻¹⁵ C/m

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