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3 years ago

14

You forgot to put what IRITONCF is unscrambled

1 answer:

Lesechka [4]3 years ago

4 0

I'm Pretty sure it's FRICTION

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A billiard ball (ball #1) moving at 5.00 m/s strikes a stationary ball (ball #2) of the same mass. after the collision, ball #1

Rashid [163]

If it is completely elastic, you can calculate the velocity of the second ball from the kinetic energy
<span>v1 = velocity of #1 </span>
<span>v1' = velocity of #1 after collision </span>
<span>v2' = velocity of #2 after collision. </span>

<span>kinetic energy: v1^2 = v1' ^2 + v2' ^2 (1/2 and m cancel out) </span>
<span>5^2 = 4.35^2 + v2' ^2 </span>
<span>v2 = 2.46 m/s <--- ANSWER</span>

4 0

3 years ago

Read 2 more answers

As longitudional waves travel, particles in the medium are pushed together and then pulled apart. We call this

andreyandreev [35.5K]

Compression and rarefraction, the other guy's answer it's wrong

4 0

3 years ago

In the deep ocean, a water wave with wavelength 95 m travels at 12 m/s. Suppose a small boat is at the crest of this wave, 1.2 m

dimaraw [331]

Answer:

the vertical position is 1.1971m

Explanation:

Recall that

$y = 1.2 * cos (\frac{2*\pi * distance travelled}{wavelenght})$

recall that $v = \frac{distance}{time}$

$thus; distance = v* t$

this implies that distance = 12 * 5.0

= 60

therefore; $y = 1.2 cos \frac{2*3.142*60}{95}$

y = 1.1971m

the

3 0

3 years ago

Sam and his Dad went grocery shopping. As they went up and down the aisle, adding items to their cart, something changed. The ca

Amanda [17]

B)
The speed of the cart changed because it stopped.

Hope I could help!
-Marshy

8 0

3 years ago

Read 2 more answers

Consider a sound wave moving through the air modeled with the equation s(x, t) = 5.00 nm cos(60.00 m−1x − 18.00 ✕ 103 s−1t). Wha

GaryK [48]

Answer:

Shortest time = 58.18 × 10^(-6) s

Explanation:

We are given;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t))

Let us set x = 0 as origin.

Now, for us to find the time difference, we need to solve 2 equations which are;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t1))

And

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t2))

Now, since the wave starts from maxima at time at t = 0, the required time would be the difference (t2 - t1)

Thus, the solutions are;

t1 = (1/(18 × 10³)) cos^(-1) (2.5/5)

And

t2 = (1/(18 × 10³)) cos^(-1) (-2.5/5)

Angle of the cos function is in radians, thus;

t1 = 58.18 × 10^(-6) s

t2 = 116.36 × 10^(-6) s

So,

Required time = t2 - t1 = (116.36 × 10^(-6) s) - (58.18 × 10^(-6) s) = 58.18 × 10^(-6) s

4 0

3 years ago