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Kitty [74]
3 years ago
13

Classify the type of plate boundary where the appalachian mountains formed . how have they changed since their formation

Physics
1 answer:
disa [49]3 years ago
6 0
<span>The Appalachian Mountains were formed when colliding tectonic plates folded and upthrust, mainly during the Permian Period and again in the Cretaceous Period. The folds and thrusts were then eroded and carved by wind, streams and glaciers. These erosive processes are ongoing, and the topography of the Appalachian Mountains continue to change. They have changed with the miles of land that are cleared of all vegetation and topsoil. In the 1970's coal miners literally blow away the top of a mountain to get to the coal underneath.</span>
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A ball is moving with velocity 5 m/s in a direction which makes an angle of 30° with horizontal (i.e. with positive x-direction)
klasskru [66]

Answer:

Explanation:

(a)

From the given information:

The initial velocity v_1 = 5 m/s

The direction of the angle θ = 30°

Therefore, the component along the x-axis = v_1 \ cos \ \theta

v_{1 \ x } = 5 \ cos \ 30^0

v_{1x} = 4.33 \ m/s

The component along the y-axis = v_2 { \ sin \  \theta}

v_{1 \ y } = 5 \ sin \ 30^0

v_{1 \ y } = 2.5 \ m/s

To find the final velocity( reflected velocity)

using the same magnitude v_2 = 5 \ m/s

The angle from the x-axis can be \theta_r = 90^0+60^0

= 150°

Thus, the component along the x-axis = v_2  \ cos \theta _r

v_{2x} = - 0.433 \ m/s

The component along the y-axis = v_2 \ sin \theta_r

v_{2y} = 5 \ sin \ 150^0

v_{2y} = 2.5 \ m/s

(b)

The velocity v_1 can be written as in vector form.

v_1 ^{\to}  = v_1 x \hat {i} + v_1 y \hat {j}

v_1 ^{\to}  =4.33 \ \hat {i} +  2.5 \  \hat {j}  ----  (1)

The reflected velocity in vector form can be computed as:

v_2 ^{\to}  = v_2 x \hat {i} + v_2 y \hat {j}

v_2 ^{\to}  =-4.33 \ \hat {i} +  2.5 \  \hat {j}  --- (2)

The change in velocity = v_2 ^{\to} - v_1 ^{\to}

\Delta v ^{\to} = - 4.33 \hat i + 2.5 \hat j - 4.33 \hat i - 2.5 \hat j

\Delta v ^{\to} = - 8.66 \hat  { i }

(c)

The magnitude of change in velocity = | \Delta V |

| \Delta V | = 8.66 m/s

6 0
3 years ago
Find the density of a planet with a radius of 8000 m if the gravitational acceleration for the planet, gp, has the same magnitud
Naya [18.7K]

Answer:

Density = 3 x 10⁻⁵ kg/m³

Explanation:

First, we will find the volume of the planet:

V = \frac{4}{3}\pi r^3\ (radius\ of\ sphere)\\\\V =   \frac{4}{3}\pi (8000\ m)^3\\\\V = 2.14\ x\ 10^{12}\ m^3

Now, we will use the expression for gravitational force to find the mass of the planet:

g = \frac{Gm}{r^2}\\\\m = \frac{gr^2}{G}

where,

m = mass = ?

g = acceleration due to gravity = 6.67 x 10⁻¹¹ m/s²

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

r = radius = 8000 m

Therefore,

m = \frac{(6.67\ x\ 10^{-11}\ m/s^2)(8000\ m)^2}{6.67\ x\ 10^{-11}\ Nm^/kg^2}\\\\m = 6.4\ x\ 10^7\ kg

Therefore, the density will be:

Density = \frac{m}{V} = \frac{6.4\ x\ 10^7\ kg}{2.14\ x\ 10^{12}\ m^3}

<u>Density = 3 x 10⁻⁵ kg/m³</u>

4 0
3 years ago
A substance has a boiling point of 220 degrees Celsius and a melting point of −120 degrees Celsius. What state will it be at roo
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Room temperature is 36 degrees celsius
7 0
3 years ago
How is a cold air mass similar to a mountain range as to the effect on a warm air mass?
Fittoniya [83]
Different types of fronts produce different patterns of weather. When a cold, dense air mass pushes warmer air, it produces a cold front. A cold front can turn a heat wave into normal summer weather or turn cold winter air into very cold weather. Cold fronts often produce tall cumulonimbus clouds.
8 0
3 years ago
What is the highest degrees above the horizon the moon ever gets during the year in the Yakima Valley ?
Ivahew [28]

The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.

Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima.  The latitude in the middle
of that intersection is 46.585° North.  <u>That's</u> the number we need.

Here's how I would do it:

-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.

-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°. 

-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.

This sets the limit of the highest in the sky that the moon can ever appear.

90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .

That doesn't happen regularly.  It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).

Depending on the time of year, that can be any time of the day or night.

The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.

In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky.  Then it's going to be somewhere near
67° above the horizon at midnight.


5 0
3 years ago
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