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3 years ago

10

3. Ang Pacific Ring of Fire ay ang malawak na sonang nakalatag sa Asya Pasipiko nanagtataglay ng mga aktibong bulkan. Alin sa mg

a bansa ang hindikabilang?
a. Pilipinas
b. China
c.Japan
d. Indonesia

Help please I would Appreciate if you do!

A.P po yan

1 answer:

Sedaia [141]3 years ago

6 0

It’s b. China.........

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Physical Science - 02.05 - Question #3

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Answer:

1. They should not trust this study. 2. Pseudoscience.

Explanation:

Because they drank the water and after 24 hours, they said that it was gone. Head aches go away sooner than that. Another reason is that some people could have lied about having a headache just so they could have the water or their headache could have gone away before they drank the water.

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3 years ago

uppose a wheel is initially rotating at 10.0 rad/s while undergoing constant angular acceleration reaching a speed of 30.0 rad/s

Paul [167]

Answer:

t = 22.2 s

Explanation:

angular distance covered in the 36.0 s is

θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians

720/2 = 360 radians

α = Δω/t = (30 - 10)/36 = 5/9 rad/s²

θ = ω₀t + ½αt²

360 = 10.0t + ½(5/9)t²

0 = (5/18)t² + 10.0t - 360

0 = t² + 36t - 1296

t = (-36 ±√(36² - 4(1)(-1296))) / 2

t = (-36 ±√(6480)) / 2

t = -18 ±√1620

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3 years ago

A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

3 0

3 years ago