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3 years ago

Why would a pencil appear to be bent if half of it was embedded in diamond and the other half embedded in glass?

1 answer:

7 0

Due to refraction. refraction is in which the fact of light, radio waves, etc.,. also the measurment of your eye charchteristics. you can always research more:)

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The diagram below shows the oscillation of a

galben [10]

Answer:

I think option B 0.125

Explanation:

but i am not sure so don't mind

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3 years ago

During which stage of the water cycle does water from the ocean form clouds?

Nataly [62]

Answer:

Condensation

Explanation:

Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.

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4 years ago

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Hemoglobin is a molecule found in blood used to transport oxygen throughout the body. What type of biological compound is this?

Hatshy [7]

Answer:

Hemoglobin is a protein (because it's a molecule that's not a living part of the cell but aids in cell processes).

8 0

3 years ago

A girl holds an arrow in her hands, ready to shoot the arrow at a dragon who is just outside her arrow’s range of 45 meters. The

gregori [183]

First, assume that the arrow leaves the bow with a velocity of 45 m/s above the horizontal with respect to the bow.

Since the bow is moving at 8.5 m/s 45º below the horizontal, find the initial velocity of the arrow with respect to the ground:

$\vec{v}_{AG}=\vec{v}_{AB}+\vec{v}_{BG}$

This equation reads:

The velocity of the arrow with respect to the ground is equal to the velocity of the arrow with respect to the bow plus the velocity of the bow with respect to the gound.

Notice that this is a vector equation. Then, the vertical and horizontal components of the velocities must be added separately:

$\begin{gathered} v_{AG-x}=v_{AB-x}+v_{BG-x} \\ v_{AG-y}=v_{AB-y}+v_{BG-y} \end{gathered}$

Find the vertical and horizontal components of the velocity of the arrow with respect to the bow and the velocity of the bow with respect to the ground:

$\begin{gathered} v_{AB-x}=v_{AB}\cos (\theta) \\ =45\frac{m}{s}\cdot\cos (30º) \\ =38.97\frac{m}{s} \end{gathered}$

$\begin{gathered} v_{AB-y}=v_{AB}\sin (\theta) \\ =45\frac{m}{s}\sin (30º) \\ =22.5\frac{m}{s} \end{gathered}$

Similarly, for the velocity of the bow with respect to the ground:

$\begin{gathered} v_{BG-x}=6.01\frac{m}{s} \\ v_{BG-y}=-6.01\frac{m}{s} \end{gathered}$

Then, the vertical and horizontal components of the initial velocity of the arrow with respect to the ground, are:

$\begin{gathered} v_{AG-x}=38.97\frac{m}{s}+6.01\frac{m}{s}=44.98\frac{m}{s} \\ \\ v_{AG-y}=22.5\frac{m}{s}-6.01\frac{m}{s}=16.49\frac{m}{s} \end{gathered}$

Use the horizontal component of the velocity to find how long it takes for the arrow to travel a horizontal distance x of 55 meters. Then, use that time to find the vertical position of the arrow.

Since the horizontal movement of the arrow is uniform, then:

$v_{AG-x}=\frac{x}{t}_{}$

Isolate t and substitute x=55m, v_{AG-x}=44.98 m/s:

$\begin{gathered} t=\frac{x}{v_{AG-x}} \\ =\frac{55m}{44.98\frac{m}{s}} \\ =1.2227s \end{gathered}$

The vertical motion of the arrow is a uniformly accelerated motion. Then, the vertical position is given by:

$y=v_{AG-y}t-\frac{1}{2}gt^2$

Replace v_{AG-y}=16.49 m/s, t=1.2227s and g=9.81 m/s^2 to find the vertical position of the arrow when the horizontal position is 55 meters. This matches the elevation of the dragon with respect to the girl when the girl shoots: