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2 years ago

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A body of mass 250g is release from the top of a building. if the body hits the ground with a velocity of 4m/s, calculate th hei

ght of the building [g=10m/s²]

1 answer:

Nutka1998 [239]2 years ago

5 0

$\purple{\maltese}\large\underline{\underline{\sf\:\: Given :}} \\$

» Mass ( m ) of the body = 250 g

» final velocity of the body = 4 m/s

$\\\purple{\maltese}\large\underline{\underline{\sf\:\: To \: \: Find :}} \\$

» The height of the Building= ??

$\\\purple{\maltese}\large\underline{\underline{\sf\:\: Solution :}} \\ \\$

★ <u>Height of the </u><u>building </u><u>by the principal of conservatio</u>n;

$\\\\\begin{gathered}\purple{\maltese}\large\underline{\underline{\sf\:\: using \: formula :}} \\ \\\end{gathered}$

$\bigstar \: \underline{ \boxed{\sf { \pink{ \: \frac{1}{2} \: \: mv { }^{2} = \: mgh \: }}}}$

$\\ \: \: \underline{\textsf {Putting values \: in \: the \: \: given \: formula :}} \\$

$\\ \sf \implies \: \frac{1}{2} \: \: mv { }^{2} = \: mgh \: \\$

$\\ \sf \implies \: \frac{v {}^{2} }{2} \: \: = \: gh \: \\$

$\\ \sf \implies \: h \: = \frac{v {}^{2} }{2g} \: \: \: \\$

$\\ \sf \implies \: h \: = \frac{16 }{20} \: \: \: \\$

$\\ \sf \implies \: h \: = 0.8 \: \: \: m \\$

$\\ \sf \implies \: h \: = 80 \: \: \: cm \\$

henceforth , The height of the Building is <u>0.8 m or 80 cm</u> ..!!!

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In this exercise, you're required to identify the action-reaction forces in the attached picture.

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The force being exerted by the wall on the boy is a force of reaction.

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The force being exerted by the feet of the boy on the diving board is a force of action.

The force being exerted by the diving board on the feet of the boy is a force of reaction.

Under condition C, a nail is being hammered into the wall.

The force being exerted by the hammer on the nail is a force of action.

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The force being exerted by the foot of the man on the ground (floor) is a force of action.

The force being exerted by the ground (floor) on the foot of the man is a force of reaction.

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Read more: brainly.com/question/15170643

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3 years ago

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Marina86 [1]

Answer:

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3 years ago

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The tape is now wound back into the drum at angular rate omega(t). With what speed will the end of the tape move? (Note that our

pshichka [43]

Answer:

See description

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The relationship is:

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Where $S$ is the arc length or distance, theta is the angle that results from the initial point of the measure to the final point, and r is the radius of a circumference.

Now let $x(t)$ be length the unwinded tape. Change $S$ by $x(t)$ and you ge the relationship:

$x(t) = \theta r$

if you unwind the tape by one revolution ( $\theta = 2 \pi$ ) you get the perimeter of a cricle $x(t)= 2 \pi r$ , if you unwind it two times then $x(t)= 4 \pi r$ and so on.

Then we have that the derivative of $x(t)$ is $v(t)$

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A small difference between means may not be statistically significant, but it could reach statistical significance with a large

aliya0001 [1]

Answer:

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The p-value can be made smaller by either

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