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Airida [17]
2 years ago
12

Which is the correct definition for sea floor spreading

Physics
2 answers:
jok3333 [9.3K]2 years ago
6 0
Seafloor spreading is a geologic process in which tectonic plates—large slabs of Earth's lithosphere—split apart from each other.
Crank2 years ago
3 0
In sea-floor spreading, the sea floor spreads apart along both sides of a mid-ocean ridge as new crust is added.


In result, the ocean floors move like conveyor belts, carrying the continents along with them
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Help. Science is getting on my nerves
anyanavicka [17]
The correct answer is the First choice
8 0
2 years ago
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Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o
Juliette [100K]

(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

U=-\frac{GMm}{R+h}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

U_A=-\frac{GMm}{R+h_A}

while potential energy of satellite B is

U_B=-\frac{GMm}{R+h_B}

Therefore the ratio of the potential energy of satellite B to that of satellite A is

\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

K=\frac{GMm}{2(R+h)}

So the kinetic energy of satellite A is

K_A=\frac{GMm}{2(R+h_A)}

while kinetic energy of satellite B is

K_B=\frac{GMm}{2(R+h_B)}

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}

which is identical to before, so it  gives

\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473

(c) Satellite B

The total energy of a satellite in orbit is given by

E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) 1.03\cdot 10^7 J

We have to calculate the total energy of each satellite.

Given:

G=6.67\cdot 10^{-11}

M=5.98\cdot 10^{24} kg

m = 12.0 kg

R+h_A = 6370 km+5920 km=12290 km = 12.3 \cdot 10^6 m

R+h_B = 6370 km+19600 km=25970 km = 26.0 \cdot 10^6 m

We find:

E_A = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(12.3\cdot 10^6)}=-1.95\cdot 10^{7} J

E_B = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(26.0\cdot 10^6)}=-9.2\cdot 10^{6} J

So the difference in total energy is

E_B-E_A = -9.2\cdot 10^6 - (-1.95\cdot 10^7) =1.03\cdot 10^7 J

6 0
3 years ago
The speed of a wave with a frequency of 2 Hz (2/s), an amplitude of 3 m, and a wavelength of 10 m is
Inessa05 [86]

Answer:

5 m/s

Explanation:

your welcome

6 0
2 years ago
In the simulation, both elastic and inelastic collisions take the same amount of time. In which type of collision is the acceler
Verdich [7]

Answer:

its ealstic.

Explanation:

5 0
3 years ago
A car with mass m travels over a hill with a radius of curvature of r at a speed of 15 m/s. What is the normal force on the car
mihalych1998 [28]

Answer:

zero

Explanation:

The computation of the normal force is shown below:

As we know that

F_c = mg - N

F_c = mv^2 ÷ r

N = mg - mv^2 ÷ r

N = m(g - v^2 ÷ r)

Assume that

The mass of the car is 1200 kg

And, r = 10 m

So,

= 1200 (9.8 - 15^2 ÷ 10)

= -15240 N

Since it comes in negative so the normal force would be zero

5 0
3 years ago
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