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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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the correct answers are 54N and -1,2m/s^2

6 0
3 years ago
Read 2 more answers
gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the
igomit [66]

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

y = \frac{\lambda L}{d}\\\\d  =\frac{\lambda L}{y}\\\\

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\

<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>

7 0
3 years ago
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms a
lawyer [7]

Answer:

This is because The energies of atoms are quantized.

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed

5 0
3 years ago
a cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5m/s2 for 8.0
Andreas93 [3]

The final velocity is +15.0 m/s

Explanation:

The motion of the cart is a uniformly accelerated motion (=at constant acceleration), therefore we can use the following suvat equation:

v=u+at

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

t is the time

For the cart in this problem, we have:

u = +3.0 m/s (initial velocity)

a=1.5 m/s^2 (acceleration)

t = 8.0 s (time)

Substituting, we find the final velocity:

v=3.0+(1.5)(8.0)=+15.0 m/s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0
3 years ago
The brakes exert a 1,951 N force on a car weighing 18,985 N and moving at 12 m/s. The car finally stops. How long (in seconds) d
nata0808 [166]

Answer:20,000

Explanation:

7 0
4 years ago
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