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3 years ago

What is the name of the theory that led to the development of the theory of plate tectonics

Physics

2 answers:

STALIN [3.7K]3 years ago

7 0

Answer:

Continental Drift

Explanation:

Continental drift is the name of the theory that led to the development of theory of plate tectonics. Theory of plate tectonics is basically latest version of Continental drift, a theory which was initially put forward by scientist “Alfred Wegener” in year 1912. But he was unable to to explain the phenomenon of movement of continents around the planet.

Plate tectonics unified all the geologic descriptions & it proposed that one ought to explain all geologic explanations as though driven by the relative motion of these tectonic plates.

noname [10]3 years ago

6 0

Answer:

continental drift

Explanation:

Continental drift is the theory that the Earth's continents have moved over geologic time relative to each other, thus appearing to have "drifted" across the ocean bed.[

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How do i find acceleration due to gravitational force?

Alexxandr [17]

Answer:

a = 9.8 m/s²

Explanation:

Acceleration due to gravity on Earth is constant, which is 9.8 m/s²

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3 years ago

3. A cat walks 0.220km North, then 0. 120 km South in a time of 400 seconds. whats the displacement and average velocity?

Andru [333]

Answer:

The rate at which velocity changes with respect to a change in time is called. acceleration.

Explanation:

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3 years ago

Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a

Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

<em>consider the attached diagram below</em>

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

$Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }$

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

$Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

Sum of the potential at point p is

V = Va + Vb

that is

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]$

the expression below can be written as the equivalent

$\frac{1}{(a^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }$

likewise,

$\frac{1}{(b^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }$

hence,

$V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

$V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

by reciprocal rule

1/a² = a⁻²

$V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]$

by binomial expansion of fractional powers

where $(1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...$

if we expand the expression we have the equivalent as shown

${(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )$

also,

${(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )$

the above equation becomes

$V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]$

$V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * \frac{1}{2r^{2} } *(b^{2} -a^{2} )$

$V = \frac{q}{8\pi e0 r^{3} } * (b^{2} -a^{2} )$

Answer

$V = \frac{q (b^{2} -a^{2} )}{8\pi e0 r^{3} }$

$V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }$

8 0

3 years ago

A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take f

sergejj [24]

Answer:A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take for the spitball to hit the ground?

(ignore air resistance) (include units and correct number of significant figures)

Explanation:La respuesta es porque esa es la respuesta, la respuesta al número es 9.7 1.8 Divide =53.888

3 0

3 years ago

If a farsighted person has a near point that is 0.600 mm from the eye, what is the focal length f2f2f_2 of the contact lenses th

Readme [11.4K]

Answer:

0.22mm

Explanation:

A far sighted person is a person suffering from long sightedness i.e such individual can only see far distant object clearly but not near distant object. The defect is corrected using convex lens.

Since convex lens is used, the focal (f) length of the lens is positive and the image distance (v) is also positive.

Using the lens formula,

1/f = 1/u + 1/v

Where u is the object distance = 0.35mm

v = 0.6mm

1/f = 1/0.35+1/0.6

1/f = 2.86 + 1.67

1/f = 4.53

f = 1/4.53

f = 0.22mm

The focal length of the contact lenses will be 0.22mm

5 0

3 years ago