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3 years ago

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Were the continents once joined together as a supercontinent? Give 3 pieces of evidence to support Alfred Wegeners Theory of Con

tinental Drift.

2 answers:

exis [7]3 years ago

8 0

Answer:

I dont know tbh

Explanation:

sattari [20]3 years ago

5 0

Yes! Fossils, The outlines of the continents and geological features .

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3.7. A dog searching for a bone walks 3.50 m south, then 8.20 m at an angle 23.1 degrees north of east, and finally 15.0 m west.

andriy [413]

Answer:

Explanation:

We shall represent displacement of dog in vector form , in terms of i , j , i representing east and j representing north .

Dog travels 3.5 m south .

Displacement D₁ = - 3.5 j

then dog travels 8.2 m , 23.1 degree north of east

Displacement D₂ = 8.2 cos23.1 i + 8.2 sin23 j

D₂ = 8.2 cos23.1 i + 8.2 sin23.1 j

= 7.54 i + 3.22 j

Third displacement

D₃ = - 15i

Total displacement = D₁ + D₂ + D₃

= - 3.5 j + 7.54 i + 3.22 j -15i

= - 7.46 i - 0.28 j

Magnitude of displacement = √ ( 7.46² + .28²)

= √(55.65 + .08 )

= 7.46 m

b ) Direction of displacement

If Ф be angle , displacement makes with west direction

TanФ = .08 / 55.65 = .00143

Ф = .082 degree south of west or almost west .

From east , this angle = 180 + .082 = 180.082 , counterclockwise .

5 0

2 years ago

If the distance between the center of two objects is quadrupled. The gravitational

Juliette [100K]

Answer:

F' = F/16

Explanation:

The gravitational force between masses is given by :

$F=G\dfrac{m_1m_2}{r^2}$

If the distance between the center of two objects is quadrupled, r' = 4r

New force will be :

$F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(4r)^2}\\\\F'=\dfrac{Gm_1m_2}{16r^2}\\\\F'=\dfrac{1}{16}\times \dfrac{Gm_1m_2}{r^2}\\\\F'=\dfrac{F}{16}$

So, the new force will change by a factor of 16.

6 0

3 years ago

What is an unbalanced force

Butoxors [25]

There's no such thing as "an unbalanced force".

If all of the forces acting on an object all add up to zero, then we say that
<span>the group </span>of forces is balanced. When that happens, the group of forces
has the same effect on the object as if there were no forces on it at all.

An example:
Two people with exactly equal strength are having a tug-of-war. They pull
with equal force in opposite directions. Each person is sweating and straining,
grunting and groaning, and exerting tremendous force. But their forces add up
to zero, and the rope goes nowhere. The <u>group</u> of forces on the rope is balanced.

On the other hand, if one of the offensive linemen is pulling on one end of
the rope, and one of the cheerleaders is pulling on the other end, then their
forces don't add up to zero, because even though they're opposite, they're
not equal. The <u>group</u> of forces is <u>unbalanced</u>, and the rope moves.

A group of forces is either balanced or unbalanced. A single force isn't.

7 0

3 years ago

A pitcher throws a baseball at 45 m/s. The baseball has a mass of 400 grams. Disregarding air resistance, the baseball's momentu

enyata [817]

So it would be 8.90 m/s kg*m/s

5 0

3 years ago

A 1.5-kg mass attached to an ideal massless spring with a spring constant of 20.0 N/m oscillates on a horizontal, frictionless t

Molodets [167]

Answer:

Explanation:

a ) angular frequency ω = $\sqrt{\frac{k}{m} }$

k is spring constant and m is mass attached

ω = $\sqrt{\frac{20}{1.5} }$

= 3.6515 rad / s

frequency of oscillation n = 3.6515 / (2 x 3.14)

= .5814 s⁻¹

x = .1 mcos(ωt)

= .1 mcos(3.6515t)

b ) maximum speed = ωA , A is amplitude

= 3.6515 x .1

= .36515 m /s

36.515 cm /s

maximum acceleration = ω²A

= 3.6515² x .1

= 1.333 m / s²

c ) Kinetic energy at displacement x

= 1/2 m ω²( A²-x²)

potential energy =1/2 m ω²x²

so 1/2 m ω²( A²-x²) = 1/2 m ω²x²

A²-x² = x²

2x² = A²

x = A / √2

6 0

3 years ago