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2 years ago

7

What is the critical angle for a ray trying to pass from glass (n = 1.52) into water (n = 1.33)? (Water n = 1.33, Air n = 1.00)

(Unit = deg)

2 answers:

ivann1987 [24]2 years ago

7 0

Answer:

61.05

Explanation:

correct on acellus

horrorfan [7]2 years ago

3 0

Answer:

59.52

Explanation:

59.5 in sigfigs

got it right on Acellus

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2 years ago

¿Hacia dónde se moverá el burro que se quiere sacar del corral jalándolo dos personas? La primer persona jala con 8 unidades de

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2 years ago

How much weight can a man lift in the jupiter if he can lift 100kg on the earth.calculate

Nuetrik [128]

Answer:

<h3><em>2</em><em>4</em><em>7</em><em>9</em><em> </em><em>Newton</em></h3>

<em>Sol</em><em>ution</em><em>,</em>

<em>Mass</em><em>=</em><em>1</em><em>0</em><em>0</em><em> </em><em>kg</em>

<em>Accele</em><em>ration</em><em> </em><em>due</em><em> </em><em>to</em><em> </em><em>gravity</em><em>(</em><em>g</em><em>)</em><em>=</em><em>2</em><em>4</em><em>.</em><em>7</em><em>9</em><em> </em><em>m</em><em>/</em><em>s^</em><em>2</em>

<em>Now</em><em>,</em><em>.</em>

<em> $weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton$ </em>

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em>

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3 years ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

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3 years ago