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3 years ago

Why does it take the water in a swimming pool some time to heat up during a hot day?

1 answer:

5 0

Because the top layer of a pool will be warmer than the bottom layer, that why filtration is important to cycle the water evenly.

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C i think & mark as brain list?

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2 years ago

4. A group of students made a rocket and launched it vertically upwards with velocity

Aleks04 [339]

Answer:

Approximately $72.9\; \rm m$ , assuming that the rocket had no propulsion onboard, and that air resistance on the rocket is negligible.

Explanation:

Initial velocity of this rocket: $u = 27\; \rm m\cdot s^{-1}$ .

When the rocket is at its maximum height, the velocity of the rocket would be equal to $0$ . That is: $v = 0\; \rm m \cdot s^{-1}$ .

The acceleration of the rocket (because of gravity) is constantly downwards, with a value of $a = -g = -10\; \rm m \cdot s^{-2}$ .

Let $x$ denote the distance that the rocket travelled from the launch site to the place where it attained maximum height. The following equation would relate $x \!$ to $u$ , $v$ , and $a$ :

$\displaystyle x = \frac{v^2 - u^2}{2\, a}$ .

Apply this equation to find the value of $x$ :

$\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{\left(0\; \rm m\cdot s^{-1}\right)^{2} - \left(27\; \rm m \cdot s^{-1}\right)^{2}}{2 \times 10\; \rm m \cdot s^{-2}} = 36.45\; \rm m\end{aligned}$ .

In other words, the maximum height that this rocket attained would be $36.45\; \rm m$ .

Again, assume that the air resistance on this rocket is negligible. The rocket would return to the ground along the same path, and would cover a total distance of $2\times 36.45\; \rm m = 72.9\; \rm m$ .

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3 years ago

When a magnet is dropped why does it lose its magnetic properties

Veronika [31]

When a magnet is dropped, the domain become unaligned.

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3 years ago

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The______ of a sound wave is defined as the amount of energy passing through a unit area of the wave front in a unit of time

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The intensity of a sound wave is defined as the amount of energy passing through a unit area of the wave front in unit of time.

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3 years ago

The distance traveled by an object can be modeled by the equation d = ut + 0.5at2 where d = distance, u = initial velocity, t =

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We are given with the expression d = ut + 0.5 at^2 and is asked to express the equation in terms of a. First, we transpose ut to the left side, then we multiply to the equation and divide lastly the resulting equation by t^2. The final expression becomes a = 2(d-ut)/t^2.

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3 years ago