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2 years ago

5

Enem 2003 embalagens longa vida brick

1 answer:

tresset_1 [31]2 years ago

3 0

Answer:

Jyftfufhfucyf fyfycyxycydyd

Explanation:

Ufivucjvk

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a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra

LenaWriter [7]

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

$a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}$

Now, we have that

${v_f}^2-{v_0}^2=2a\Delta x$

so we end up with a distance traveled of

$\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

$\implies\Delta x=79.6\,\mathrm m$

6 0

3 years ago

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let $p_{up}$ be the pressure of a point just above the wing.

Let $p_{do}$ be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

$\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}$

So, $p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa$

Force is given by the product of pressure difference and area.

Given that area is $27ms^{2}$ .

So,lifting force is $16398.48\times 27=442758.96N$

6 0

3 years ago

A particle executes simple harmonic motion with an amplitude of 2.18 cm.

Bogdan [553]

Answer:

The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.

Explanation:

Maximum speed is :

v (max) = Aω

Speed v at any displacement y is given by

$v^{2}$ = $w^{2}$ ( $A^{2}$ - $y^{2}$ ) ........................................................ i

And,

v = $\frac{1}{2}$ v (max)

or, 2 × v = Aω .................................................... ii

Eliminating ω from equations i and ii,

$\frac{1}{4}$ $A^{2}$ $w^{2}$ = $w^{2}$ ( $A^{2}$ - $y^{2}$ )

or, $y^{2}$ = ( $\frac{3}{4}$ ) $A^{2}$ =( $\frac{3}{4}$ ) $2.18^{2}$

or, y = 3.56 cm.

3 0

2 years ago

PLEASE help me quickly

Anna007 [38]

Explanation:

its either a or d however i say the best choice is d

6 0

2 years ago

What does the formula "F=m xa" mean?

strojnjashka [21]

Answer:

Force = Mass x Acceleration

Explanation:

5 0

3 years ago