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3 years ago

Evaporation only occurs at the surface of a liquid

Physics

1 answer:

Veseljchak [2.6K]3 years ago

8 0

yes

evaporation starts on the surface

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What is the scientific method

MissTica

Answer:

Step 7- Communicate. Present/share your results. Replicate.

Step 1- Question.

Step 2-Research.

Step 3-Hypothesis.

Step 4-Experiment.

Step 5-Observations.

Step 6-Results/Conclusion

Explanation:

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3 years ago

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The acceleration of a block attached to a spring is given by a=−(0.324m/s2)cos([2.50rad/s]t) a = − ( 0.324 m / s 2 ) c o s ( [ 2

allsm [11]

Answer:

Looks like you have:

a = -.324 cos 2.5 t

In this case ω^2 A = .324

ω = 2.5

f = ω / (2 * pi) = 2.5 / 6.28 = .40 / sec

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2 years ago

Pls help! True or False?

jasenka [17]

Answer:

True

Explanation:

i hope it helps

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2 years ago

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Any four difference between velocity and acceleration

pychu [463]

Answer:

https://physicsabout.com/acceleration-and-velcoity/

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3 years ago

A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c

kirill [66]

Answer:

Velocity of the car at the bottom of the slope: approximately $20.3\; \rm m \cdot s^{-2}$ .

It would take approximately $3.9\; \rm s$ for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

Let $v$ denote the final velocity of the car.
Let $u$ denote the initial velocity of the car.
Let $a$ denote the acceleration of the car.
Let $x$ denote the distance that this car travelled.

$v^2 - u^2 = 2\, a\cdot x$ .

Given:

$u = 3\; \rm m \cdot s^{-1}$ .
$a = 4.5\; \rm m \cdot s^{-2}$ .
$x = 45\; \rm m$ .

Rearrange the equation $v^2 - u^2 = 2\, a\cdot x$ and solve for $v$ :

$\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}$ .

Calculate the time required for reaching this speed from $u = 3\; \rm m \cdot s^{-1}$ at $a = 4.5\; \rm m \cdot s^{-2}$ :

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}$ .

3 0

3 years ago