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2 years ago

Where is the natural light display called aurora borealis located?

Physics

1 answer:

Xelga [282]2 years ago

5 0

The natural light display called aurora borealis is located in the northern

hemisphere.

There are two types of aurora which are called aurora borealis and aurora

australis. The aurora borealis is located in the Northern hemisphere while

the aurora australis is located in the Southern hemisphere.

They receive their energy through the interaction of charged particles

on the Sun and Earth to produce the light display. An example

of the interaction involves solar wind with atoms of the upper atmosphere.

Read more on brainly.com/question/20191244

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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:

<em>The velocity of the arrow with respect to the ground </em><em>is equal to</em><em> the velocity of the arrow with respect to the bow </em><em>plus</em><em> the velocity of the bow with respect to the gound.</em>

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:

$\begin{gathered} y=(16.49\frac{m}{s})(1.2227s)-\frac{1}{2}(9.81\frac{m}{s^2})(1.2227s)^2 \\ =12.829\ldots m \\ \approx12.8m \end{gathered}$

Therefore, the dragon is 12.8 meters above the girl when the arrow is shoot.

6 0

1 year ago

a golfer tees off and hits a golf ball at a speed of 31 m/s and an angle of 35 degrees. how far did the ball travel before hitti

poizon [28]

Ans: R = Ball Travelled = 92.15 meters.

Explanation:
First we need to derive that formula for the "range" in order to know how far the ball traveled before hitting the ground.

Along x-axis, equation would be:
$x = x_o + v_o_xt + \frac{at^2}{2}$

Since there is no acceleration along x-direction; therefore,
$x = x_o + v_o_xt$

Since $v_o_x = v_ocos \alpha$ and $x_o$ =0; therefore above equation becomes,

$x = v_ocos \alpha t$ --- (A)

Now we need to find "t", and the time is not given. In order to do so, we shall use the y-direction motion equation. Before hitting the ground y ≈ 0 and a = -g; therefore,

=> $y = y_o + v_o_yt - \frac{gt^2}{2}$
=> $t = \frac{2v_o_y}{g}$

Since $v_o_y = sin \alpha$ ; therefore above equation becomes,
$t = \frac{2v_osin \alpha }{g}$

Put the value of t in equation (A):

(A) => $x = v_ocos \alpha \frac{2v_osin \alpha }{g}$

Where x = Range = R, and $2sin \alpha cos \alpha = sin(2 \alpha )$ ; therefore above equation becomes:

=> $R = (v_o)^2 *\frac{sin(2 \alpha )}{g}$

Now, as:
$v_o = 31 m/s$

and $\alpha = 35$ °
and g = 9.8 m/(s^2)

Hence,
$R = (31)^2 *\frac{sin(2 *35 )}{9.8}$

Ans: R = 92.15 meters.

-i

7 0

3 years ago

Jocelyn is running for the U.S Senate. What is one of the qualifications for office she must meet?

VMariaS [17]

Answer:

Explanation:

4 0

2 years ago

Read 2 more answers

What is the transfer of energy by collisions between the atoms in the material

Anettt [7]

Conduction is the transfer of energy by collusion between the atoms and molecules in a material.

5 0

3 years ago

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Question 6

Margaret [11]

Limestone and dolomite are the rocks present in the locations which leads to the formation of caves.

<h2>Formation of caves</h2>

The type of rocks that once existed in these locations are limestone and dolomite whereas the pH of the nearby groundwater is slightly acidic which is responsible for the formation of caves. Caves are formed by the dissolution of limestone due to acid rain.

Rainwater reacts with carbon dioxide from the air and percolates through the soil, which turns into a weak acid. This slowly dissolves out the limestone which become turn to form caves so we can conclude that Limestone and dolomite are the rocks present in the locations which leads to the formation of caves.

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4 0

2 years ago