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IceJOKER [234]
3 years ago
12

Check and double-click on the red polygon for the Henry Mountains laccolith complex in the folder labeled Problem 1. This comple

x consists of many blister-like intrusions. Note the size difference between this intrusion and the Sierra Nevada batholith. Estimate how much smaller in length the Henry Mountains laccolith complex is relative to the batholith. Select one:a. ~1/2b. ~1/4c. ~1/6d. ~1/17
Physics
2 answers:
Kobotan [32]3 years ago
8 0

Answer:

d

Explanation:

The batholith is roughly 720km, while the laccolith is roughly 42km.  Take 42/720 and it gives you roughly 1/17/

MArishka [77]3 years ago
4 0

Answer: ghg

Explanation:

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allochka39001 [22]
B is the correct answer
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3 years ago
A 10.0g piece of copper wire, sitting in the sun reaches a temperature of 80.0 C. how many Joules are released when the copper c
Zolol [24]

Answer:

150.8 J

Explanation:

The heat released by the copper wire is given by:

Q=mC_s \Delta T

where:

m = 10.0 g is the mass of the wire

Cs = 0.377 j/(g.C) is the specific heat capacity of copper

\Delta T=40.0 C - 80.0 C=-40.0 C is the change in temperature of the wire

Substituting into the equation, we find

Q=(10.0 g)(0.377 J/gC)(-40.0^{\circ})=-150.8 J

And the sign is negative because the heat is released by the wire.

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What body parts were scientists wanting to image that prompted the development of the CT scanner
zalisa [80]

Answer:

The head

Explanation:

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What medium is often used to transfer computer information? a. Air c. Glass b. Water d. Optic fibers Please select the best answ
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7 0
3 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
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