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4 years ago

The Pangaea theory supports the theory of plate tectonics because _____.

Physics

2 answers:

shutvik [7]4 years ago

8 0

It is based on the idea that all the present continents were on supercontinent.

Valentin [98]4 years ago

8 0

Answer: your answer is B. it is based on the idea that all the present continents were on supercontinent.

Explanation:I took the test

You might be interested in

What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter

Kobotan [32]

Answer:

I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).

The moment of inertia about the center of a sphere is 2 / 5 M R^2.

By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is I = 2/5 M R^2 + M R^2 = 7/5 M R^2

I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2

7 0

3 years ago

The angular position of objects as a function of time is given, where a, b, and care constants. In which of these cases is the a

GalinKa [24]

Answer:

Explanation:

The options is not well presented

This are the options

A. θ = at³ + b

B. θ = at² + bt + c

C. θ = at² — b

D. θ = Sin(at)

So, we want to prove which of the following option have a constant angular acceleration I.e. does not depend on time

Now,

Angular acceleration can be determine using.

α = d²θ / dt²

α = θ''(t)

So, second deferential of each θ(t) will give the angular acceleration

A. θ = at³ + b

dθ/dt = 3at² + 0 = 3at²

d²θ/dt² = 6at

α = d²θ/dt² = 6at

The angular acceleration here still depend on time

B. θ = at² + bt + c

dθ/dt = 2at + b + 0 = 2at + b

d²θ/dt² = 2a + 0 = 2a

α = d²θ/dt² = 2a

Then, the angular acceleration here is constant is "a" is a constant and the angular acceleration is independent on time.

C. θ = at² —b

dθ/dt = 2at — 0 = 2at

d²θ/dt² = 2a

α = d²θ/dt² = 2a

Same as above in B. The angular acceleration here is constant is "a" is a constant and the angular acceleration is independent on time.

D. θ = Sin(at)

dθ/dt = aCos(at)

d²θ/dt² = —a²Sin(at) = —a²θ

α = d²θ/dt² = -a²θ

Since θ is not a constant, then, the angular acceleration is dependent on time and angular displacement

So,

The answer is B and C

4 0

3 years ago

Where is the center of mass of the Earth-Moon system? The radius of the Earth is 6378 km and the radius of the Moon is 1737 km.

lions [1.4K]

Answer:

r_{cm} = 4.67 10⁶ m

We see that none of the answers is correct, the closest is the third, but the center of mass is within the radius of the Earth

Explanation:

The definition of the center of mass is

$r_{cm}$ = 1 / M ∑ $r_{i} m_{i}$

Where M is the total mass, $r_{i}$ and $m_{i}$ mi are the position and the mass gives each of the bodies

We must locate a reference system for the calculation, we will locate it with the origin in the center of the Earth, the data we have are

Mass of the Earth Me = 5.98 1024 kg

Moon mass m = 7.36 1022 kg

Earth to Moon Distance r = 3.84 108 m

Let's apply this to our case

r_{cm} = 1 / (Me + m) (Me 0 + m R)

r_{cm} = 1 / (598 +7.36) 10²² (0 + 7.36 10²² 3.84 10⁸)

r_{cm} = 4.67 10⁶ m

We can see that this distance is less than the radius of the Earth

We see that none of the answers is correct, the closest is the third, but the center of mass is within the radius of the Earth

3 0

4 years ago

Can you guys help me with this guestion?

OlgaM077 [116]

Answer:

Explanation:

A -heat capacity

B - continuity equation

C -pressure

D - mechanical energy

3 0

3 years ago

At high noon, the sun delivers 1150 w to each square meter of a blacktop road. if the hot asphalt loses energy only by radiation

Pie

Stephen`s Law:
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
$T = \sqrt[4]{202.857 * 10 ^{8} }$
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K.

3 0

3 years ago