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3 years ago

8

What if the store decided to arrange their items according to size, color or shape? What do you think will happen to your shoppi

ng experience?

1 answer:

Y_Kistochka [10]3 years ago

8 0

Answer: it will change dramatically

Explanation:

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Can anyone answer my questions that I've asked?

Monica [59]

Answer:

76 N north-east

Explanation:

V = √(44²+62²) = 76 N

8 0

3 years ago

4. A ball is thrown with an initial speed vi at an angle θi with the horizontal. The

GalinKa [24]

Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time $t$ , the horizontal position $x$ and vertical position $y$ of the ball are given respectively by

$x = v_i \cos(\theta_i) t$

$y = v_i \sin(\theta_i) t - \dfrac g2 t^2$

and the horizontal velocity $v_x$ and vertical velocity $v_y$ are

$v_x = v_i \cos(\theta_i)$

$v_y = v_i \sin(\theta_i) - gt$

The ball reaches its maximum height with $v_y=0$ . At this point, the ball has zero vertical velocity. This happens when

$v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g$

which means

$y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g$

At the same time, the ball will have traveled half its horizontal range, so

$x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g$

Solve for $v_i$ and $\theta_i$ :

$\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0$

Since $0^\circ$ , we cannot have $\sin(\theta_i)=0$ , so we're left with (e)

$3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}$

Now,

$\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}$

$\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}$

so it follows that (d)

$R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}$

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

$v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}$

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

$v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}$

Then with $v_y=0$ , the ball's speed $v$ is

$v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}$

Finally, in the work leading up to part (e), we showed the time to maximum height is

$t = \dfrac{v_i \sin(\theta_i)}g$

but this is just half the total time the ball spends in the air. The total airtime is then

$2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}$

and the ball is in the air over the interval (a)

$\boxed{0 < t < 2\sqrt{\frac R{3g}}}$

7 0

2 years ago

An student dropped their book from the top of a canyon into a river 40 meters below.

kotykmax [81]

Use this formula for an object starting from rest and accelerating:

Distance = (1/2) (acceleration) (time)²

In this problem, the distance is 40 meters, and acceleration is gravity.

So . . .

40 m = (1/2) (9.8 m/s²) (time)²

Divide each side by (4.9 m/s²) :

time² = (40 m) / (4.9 m/s²)

time² = 8.16 sec²

Take the square root of each side :

√(time²) = √(8.16 sec²)

time = 2.86 seconds

Round it to 2.9 sec <em>(choice B)</em>

7 0

4 years ago

What is the average velocity in the time interval 3 to 4 seconds

AlekseyPX

Since the position doesn't change over that time, it's zero

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4 years ago

Read 2 more answers

Why are hurricanes considered more damaging than tornadoes when tornadoes have stronger winds

chubhunter [2.5K]

-- A tornado follows a path that's a few miles wide, for a few hours.
   Then it's all over.

-- A hurricane follows a path that's several hundred miles wide,
   for a week or two, before it's over.
   Then comes the rain, continuing on the same path, for another week.

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3 years ago

Read 2 more answers