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e-lub [12.9K]
2 years ago
13

I really need help can anyone help

Physics
1 answer:
sergiy2304 [10]2 years ago
5 0

Answer:

percentage w a slug preference (start) - 35% (0.35)

percentage w a fish preference (start) - 65% (0.65)

percentage w a slug preference (end) - 25% (0.25)

percentage w a fish preference (end) - 75% (0.75)

hope this helps kind stranger

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2 years ago
4. A group of students made a rocket and launched it vertically upwards with velocity
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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
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Answer:

'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Explanation:

The question is incomplete, find the complete question in the comment section.

Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved  mirror than the centre of curvature.</em>

<em></em>

During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>

Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

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3 years ago
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