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2 years ago

Can you please answer my question correctly ASAP

Physics

1 answer:

lana66690 [7]2 years ago

5 0

Answer:

1) C, 2) C, 3) B, 4) D, 5) A/B, 6) D, 7) D, 8) A, 9) B, 10) A

Explanation:

1) Seconds, minutes and days are time units. (Correct answer: C)

2) The displacement is shortest distance between the point of origin and the point of destination. (Correct answer: C)

3) The velocity is the rate of change of position in time. (Correct answer: B)

4) Graphs are a very useful ressource to infer characteristics inherent to a studied physical variable. (Correct answer: D)

5) There are two possible answers, since both milimeters and meters are units for distance. (Correct answers: A/B)

6) The distance is the arc length of the entire path that object travelled. (Answer: D)

7) Traditionally, in a position vs. time graph we plot time in the x-axis. (Answer: D)

8) In this case, we use a distance-time graph, where travelled distance is the dependent variable and time is the independent variable. (Answer: A)

9) The point of origin is the initial position of the object. (Answer: B)

10) The point of destination is the final position of the object. (Answer: A)

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3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

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Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant