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3 years ago

Which of the following is a characteristic of science?

Physics

2 answers:

vfiekz [6]3 years ago

7 0

Answer:

Using controlled conditions

Explanation:

Science experiments and tests are always carried out using specific conditions and parameters. These can be pressure, temperature, volume or the close surrounding around us which determines a 'tentative' results. The environment or the surrounding in which the tests are run plays a very important role and it has large affect on the result of the experiment. Thus these scientific experiment are carried at controlled conditions to obtain the desired results.

Thus the answer is, Using controlled conditions.

Oksi-84 [34.3K]3 years ago

6 0

Answer:

C. Using controlled conditions

Explanation:

I did the quiz and B turned out to be wrong. D just doesn't make sense. Although A can be true in some experiments, it's not always the case.

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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

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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

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