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ikadub [295]
3 years ago
9

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the

Displacement the Velocity the Acceleration None of the above
Physics
1 answer:
dolphi86 [110]3 years ago
5 0

Answer:

Please find the answer in the explanation

Explanation:

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above

The slope of any time graph can not give you distance or displacement except for position - time graph.

When you plot either distance or displacement against time, that is, distance time graph or displacement time graph, you can get speed or velocity as the slope of the line segment.

You can only acceleration as a slope in a line of best fit if velocity is plotted against time. That is, in a velocity time graph.

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Answer:

seconds (s) = hours (h) *3,600 ; h = \frac{s}{3,600} \\g = kg * 1,000; kg = \frac{g}{1,000} \\cm = \frac{m}{100};m = cm * 100

1. a) 0.5 h = 1,800 s

  h) 20 cm = 0.2 m

  b) 2.0 h = 7,200 s

   i) 5.0 kg = 5,000 g

  c) 3.5 h = 12,600 s

  j) 1.5 kg = 1,500 g

  d) 1/4 h = 900 s

  k) 450.0 g = 0.45 kg

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3 years ago
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The truck has more KE than the bike
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Hii please help i’ll give brainliest if you give a correct answer please please hurry it’s timed
Zina [86]

Answer: The answer is A for sure

Explanation:

That is, there will be no acceleration. If you are sitting at rest in a chair and the upward push of the chair is equal to the downward pull of gravity, you will stay at rest in the chair. ... You now have an unbalanced force acting on you and therefore, according to Newton's First Law, your motion is going to change.

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3 years ago
Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co
patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of ^{3890}Sr\, t_{\frac{1}{2}} = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where

N = No. of nuclei left after time t

N_{o} = No. of nuclei initially started with

\frac{N}{N_{o}} = 1\times 10^{- 4}

(Since, 100% - 99.99% = 0.01%)

Thus

1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}

Taking log on both the sides:

- 4 = \frac{t}{28.5}log\frac{1}{2}

t = \frac{-4\times 28.5}{log\frac{1}{2}}

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3 years ago
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