All categories

3 years ago

Why do astronomers use the word on to describe angles on the sky rather than angles in the sky?

1 answer:

8 0

The use of the word on instead of the word in when referring to the angular distance between celestial objects comes about because all of the objects appear to be on the celestial sphere and at an indeterminable distance. While we know that objects are at different distances in the sky, their distance from Earth is irrelevant in determining the angular distance between the two objects as viewed from Earth.

You might be interested in

If a piece of space debris is too large to be a meteoroid and too small to be a planet, it could be

Vaselesa [24]

an asteroid was the correct answer

3 0

3 years ago

Read 2 more answers

Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction

Norma-Jean [14]

Answer:

The ball would hit the floor approximately $0.55\; \rm s$ after leaving the table.

The ball would travel approximately $5.5\; \rm m$ horizontally after leaving the table.

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Let $\Delta h$ denote the change to the height of the ball. Let $t$ denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let $v_0(\text{vertical})$ denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible: $\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t$ .

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was $v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}$ .

The height of the table was $1.5\; \rm m$ . Therefore, after hitting the floor, the ball would be $1.5\; \rm m \!$ below where it was before leaving the table. Hence, $\Delta h = -1.5\;\rm m$ .

The equation becomes:

$\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}$ .

Solve for $t$ :

$\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55$ .

In other words, it would take approximately $0.55\; \rm s$ for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at $v(\text{horizontal}) =10\; \rm m \cdot s^{-1}$ ) until the ball hits the floor.

The ball was in the air for approximately $t = 0.55\; \rm s$ and would have travelled approximately $v(\text{horizontal})\cdot t \approx 5.5\;\rm m$ horizontally during the flight.

4 0

2 years ago

One Dimensional Kinematics Escape Room level 3?

Fed [463]

Sorry but we don’t know that

6 0

3 years ago

What was the "Plum Pudding" model of the atom and it's electrons?

MrMuchimi

The 'plum pudding' model of the atom was proposed by JJ Thomson, who had also discovered the electron. It was put forth before the discovery of the nucleus. According to this model, the atom is a sphere of positive charge, and negatively charged electrons are embedded in it to balance the total positive charge.

7 0

2 years ago

Read 2 more answers

If there is a base runner on 1st base, then there is a(n) _________ at 2nd base

Yuri [45]

Answer:4

Explanation:

8 0

3 years ago