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Sladkaya [172]
3 years ago
13

The best ways to determine the safest driving speed are to 1) know the speed limit, and 2) ________.

Physics
1 answer:
likoan [24]3 years ago
7 0
The best ways to determine the safest driving speed are to 1) know the speed limit, and 2) maintain a safe following distance.<span>  The safe following distance is to be enough distanced form the vehicle in front of you. But how much is enough?
A </span>driver<span> should ideally stay at least two seconds behind any vehicle that is directly in front of his or her vehicle.</span>
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In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400
liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

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Julie's psychologist helped her get over her fear of spiders. Which goal of psychology has the phycologist met ?
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<span>The four goals of psychology are: describe observed phenomena, explain them, predict what may occur as a result of them and control behaviour as a result. Julie’s psychologist has helped Julie control her behaviour - her fear response to spiders - so they have met the ‘control behaviour’ goal of psychology.</span>
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An asteroid is a minor planet of the inner Solar System. Historically, these terms have been applied to any astronomical object orbiting the Sun.

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Why does the entropy of a gas increase when it expands into a vacuum? Why does the entropy of a gas increase when it expands int
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Answer:

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Explanation:

When a gas expand in a vacuum, the molecules of the gases vibrates very fast and starting moving with higher velocity in random directions which means the level of disorder in the gases increases.

Now the possible state of the gas molecule increases such as the particle can be located at different position due to increased randomness.

<u>Entropy is the measure of this randomness and thus with this increased randomness entropy also increases.</u>

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We r made of atom so v can’t touch anything hehe I just joking
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