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3 years ago

What causes us to perceive objects in three dimensions?

Physics

1 answer:

Ymorist [56]3 years ago

5 0

Answer:

Light bouncing or reflecting off an object in different ways.

I just took a quiz and this was the answer

(k12 answer)

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Part 1

xz_007 [3.2K]

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ .

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ .

<h3>Determination of the coordinates of the center of gravity</h3>

Let suppose that each square has an <em>uniform</em> density, the coordinates of the center of gravity of each square with respect to the origin are, respectively:

$\vec r_{1} = (0.5\cdot a, 0.5\cdot a)$

$\vec r_{2} = (1.5\cdot a, 0.5\cdot a)$

$\vec r_{3} = (0.5\cdot a, 1.5\cdot a)$

$\vec r_{4} = (1.5\cdot a, 1.5\cdot a)$

The center of gravity of the <em>entire</em> system is found by applying definition of <em>weighted</em> averages:

$\vec r_{cg} = \frac{W_{1}\cdot \vec r_{1}+W_{2}\cdot \vec r_{2}+W_{3}\cdot \vec r_{3}+W_{4}\cdot \vec r_{4}}{W_{1}+W_{2}+W_{3}+W_{4}}$

Where $W_{1}$ , $W_{2}$ , $W_{3}$ and $W_{4}$ are weights of the each square, in newtons.

Now we proceed the coordinates of the center of gravity of the entire system:

$\vec r_{cg} = \frac{(50\,N)\cdot (0.5\cdot a, 0.5\cdot a) + (30\,N)\cdot (1.5\cdot a, 0.5\cdot a)+(30\,N)\cdot (0.5\cdot a, 1.5\cdot a)+(70\,N)\cdot (1.5\cdot a, 1.5\cdot a)}{50\,N+30\,N+30\,N+70\,N}$

$\vec r_{cg} = \frac{5}{18}\cdot (0.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (1.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (0.5\cdot a, 1.5\cdot a) + \frac{7}{18}\cdot (1.5\cdot a, 1.5\cdot a)$

$\vec r_{cg} = \left(\frac{19}{18}\cdot a, \frac{19}{18}\cdot a \right)$ $\blacksquare$

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ . $\blacksquare$

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ . $\blacksquare$

To learn more on center of gravity, we kindly invite to check this verified question: brainly.com/question/20662119

6 0

2 years ago

How to find time In flight

Romashka-Z-Leto [24]

To solve any problem in math or physics, you collect all the formulas,
laws, rules, and equations that you know about the subject, then, along
with the given information, use them to find the missing information.

So the answer to "how" really depends on what information you're given.

3 0

3 years ago

The muzzle velocity of a rifle bullet is 800. m⋅ s −1 m⋅s−1 along the direction of motion. If the bullet weighs 27.0 g g , and t

11Alexandr11 [23.1K]

Answer:

1.98 × 10⁻³³m

Explanation:

It is given that,

Mass of the bullet, m = 27 g = 0.027 kg

Velocity of bullet, v = 800 m/s

The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :

$p=mvp=0.027\times 800 = 21.6\ kg-m/s$

Uncertainty in momentum is,

$\Delta p=0.2\%\ of\ 21.6\\\Delta p=0.0432$

We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :

$\Delta p.\Delta x\geq \dfrac{h}{4\pi}\Delta x=\dfrac{h}{4\pi \Delta p}\Delta x=\dfrac{6.62\times 10^{-34}}{4\pi \times 0.0432}\Delta x=1.98\times 10^{-33}\ m$