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3 years ago

Mariner 10 was the first to visit this planet in 1974. what is this planet?

Physics

1 answer:

hram777 [196]3 years ago

6 0

<h2>Answer: Mercury</h2><h2 />

The Mariner 10 probe was launched by NASA on November 3rd, 1973, with the purpose of exploring the characteristics of two planets in the solar system that were closest to the Sun, Mercury and Venus.

In addition, it was launched to explore the atmosphere and surface of both planets and <u>prove that it was possible to use gravitational assistance</u> (also called <u>slingshot effect</u>, a special orbital maneuver in order to use the gravitational field energy of a planet or massive body to accelerate or slow the probe and change the direction of its trajectory) in long interplanetary trips to save fuel.

In this case, <u>Mariner 10 first arrived at Venus</u> and succeded in using its gravitational field to accelerate its trajectory towards Mercury.

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7nadin3 [17]

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3 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

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I think the answer is twice as much

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