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Snowcat [4.5K]
3 years ago
8

Volcanoes often form belts along plate boundaries. How does the piece of art illustrate that this statement holds true for North

America? explain.
Physics
1 answer:
STALIN [3.7K]3 years ago
4 0

Answer:

The formation of volcanic arcs, trenches and subsidence leads to the production of volcanic eruptions.

Explanation:

  • North America plate makes a convergent boundary along with the pacific oceans. The is man reason for the development of the volcanoes and the origin of the earthquakes. The collusion of both these pales leads to the formation of a volcanic ring of fire that can be seen in the pacific ocean.
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Calculate the force that the 4kg block exerts on the 10kg block
Kryger [21]
Acceleration=force/mass=28/(10+4)=2m/s^2

force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block

vf=vi+at
-10=20*28/14 * t
t=30/2=15sec

i hope this can help you.
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3 years ago
Which statement about oceans is incorrect?
Hitman42 [59]
It is C because less than one percent of water is fresh water
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Neutron stars consist only of neutrons and have unbelievably high densities. a typical mass and radius for a neutron star might
Tanya [424]
<span>Density is 3.4x10^18 kg/m^3 Dime weighs 1.5x10^12 pounds The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so 4/3 pi 1.9x10^3 = 4/3 pi 6.859x10^3 m^3 = 2.873x10^10 m^3 Now divide the mass by the volume 9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3 Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3 Now to figure out how much the dime weighs, just multiply by the volume of the dime. 3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg And to convert from kg to lbs, multiply by 2.20462, so 6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>
4 0
3 years ago
Need some help with this!
zvonat [6]

Answer: what’s it asking

Explanation:

8 0
3 years ago
Read 2 more answers
An electron in a cathode-ray beam passes between 2.5cm long parallel-plate electrodes that are 6.0mm apart. A 2.1mT, 2.5-cm-wide
Dmitry_Shevchenko [17]

Answer:

(a). The speed of electron is 1.56\times10^{7}\ m/s.

(b). The radius of electron is 4.2\ cm

Explanation:

Given that,

Length = 2.5 cm

Distance = 6.0 mm

Magnetic field = 2.1 T

Potential difference = 700 V

(a). We need to calculate the electron's speed

Using formula of speed

v=\sqrt{\dfrac{2eV}{m}}

Put the value into the formula

v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}

v=15689290.81\ m/s

v=1.56\times10^{7}\ m/s

(b). We need to calculate the radius of electron

Using formula of centripetal force

\dfrac{mv^2}{r}=qvB

r=\dfrac{mv}{qB}

Where,

m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}

r=0.042\ m

r=4.2\ cm

Hence, (a). The speed of electron is 1.56\times10^{7}\ m/s.

(b). The radius of electron is 4.2 cm

8 0
3 years ago
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