Bill runs 400 meters to Andy's house, turns around, and runs 400 meters back home. What is Bill's displacement? A. 1600 meters B. 400 meters C. 0 meters D. 800 meters​

Answers

Answer 1

Answer:

C - 0 meters

Explanation:

Displacement is the difference in where you started and where you end up. If Bill starts at his house and ends at his house, there was no change in position, meaning Bill had no displacement. In math this would be

400 + (-400) = 400 - 400 = 0

assuming he runs in the positive direction and then in the negative direction. Either way you put it, it still ends up as 0


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LOL

God Bless brother. I hope you get it. It takes hard work.

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the 20 kg at angle of 53⁰ in an inclined plane is realsed from rest the coefficient of friction bn the block and the inclined plane are ųs=0.3 and ųk=0.2
A) would the block move
B) if it moves what is its speed after it has descended a distance of 5m down the plane​

Answers

Given :-

A 20kg block at an angle 53⁰ in an inclined plane is released from rest .[tex]\mu_s = 0.3 \ \& \ \mu_k = 0.2[/tex]

To Find :-

Would the block move ?If it moves what is its speed after it has descended a distance of 5m down the plane .

Solution :-

For figure refer to attachment .

So the block will move if the angle of the inclined plane is greater than the angle of repose . We can find it as ,

[tex]\longrightarrow \theta_{repose}= tan^{-1}(\mu_s)[/tex]

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[tex] \longrightarrow \theta_{repose}= tan^{-1}( 0.2)[/tex]

Solve ,

[tex]\longrightarrow\underline{\underline{\theta_{repose}= 16.6^o }}[/tex]

Hence ,

[tex]\longrightarrow\theta_{plane}>\theta_{repose}[/tex]

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[tex]\longrightarrow F_n = mgsin53^o - \mu_k N [/tex]

Substitute, N = mgcos53⁰ ( see attachment)

[tex] \longrightarrow ma_n = mgsin53^o - \mu_k mgcos53^o [/tex]

Take m as common,

[tex] \longrightarrow\cancel{m }(a_n) = \cancel{m}( gsin53^o - \mu gcos53^o)[/tex]

Simplify ,

[tex] \longrightarrow a_n = gsin53^o - \mu_k g cos53^o[/tex]

Substitute the values of sin , cos and g ,

[tex] \longrightarrow a_n = 10( 0.79 - 0.2 (0.6)) [/tex]

Simplify ,

[tex] \longrightarrow a_n = 10 ( 0.79 - 0.12 ) \\\\ \longrightarrow a_n = 10 (0.67)\\\\ \longrightarrow \underline{\underline{a_n = 6.7 m/s^2}}[/tex]

Now using the Third equation of motion namely,

[tex] \longrightarrow2as = v^2-u^2[/tex]

Substituting the respective values,

[tex] \longrightarrow2(6.7)(5) = v^2-(0)^2 [/tex]

Simplify and solve for v ,

[tex] \longrightarrow v^2 = 67 m/s\\\\\longrightarrow v =\sqrt{67} m/s \\\\\longrightarrow\underline{\underline{ v = 8.18 m/s }} [/tex]

Hence the velocity after covering 5m is 8.18 m/s .

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Answers

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Explanation:

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Answer:

Explanation:

Solution,

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A.speed=60/5=12m/s

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work and force

w = f cos 23 . d

w = 25 cos 23° . 18

w = 414.23 J

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