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pogonyaev
3 years ago
5

Which best describes the ages of the rocks on opposite sides of and the same distance from, A Seafloor spreading center?

Physics
1 answer:
Genrish500 [490]3 years ago
4 0

Answer;

-The rocks are the same age

Explanation;

Seafloor spreading is the process by which the seafloor moves apart at mid-ocean ridges.  Divergent seafloor spreading occurs at this type of plate boundary.

Seafloor spreading and other tectonic activity processes are the result of mantle convection. Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle’s convection currents makes the crust more plastic and less dense.


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Which statement does not describe the relationship between genes and
jek_recluse [69]

Answer: C. Inherited traits carry the instructions for individual genes.

Explanation: bejewels I know stuff. ☆ - ~ hope this helps

4 0
3 years ago
Read 2 more answers
In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to
goldenfox [79]

Answer:

17304 J

Explanation:

Complete statement of the question is :

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 84 kg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N .

Part A

How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

Solution :

T = tension force in the tow rope = 350 N

L  = length of the incline surface = 120 m

W_{t}  = work done by tension force = ?

The tension force acts parallel to incline surface, hence work done by tension force is given as

W_{t} = T L\\W_{t} = (350) (120)\\W_{t} = 42000 J

h = height gained by the rider = 30 m

m = total mass of rider and tube = 84 kg

Potential energy gained is given as

U = mgh\\U = (84) (9.8) (30)\\U = 24696 J

Q  = Thermal energy created

Using conservation of energy

Q = W_{t} - U\\Q = 42000 - 24696\\Q = 17304 J

7 0
3 years ago
the chef was careless and put the aluminum cookie sheet directly on the hot stove, which melted 0.05 kg of the aluminum. how muc
sergejj [24]

Heat required to melt 0.05 kg of aluminum is 28.7 kJ.

<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>

The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.

The formula to be used is given below:

  • Heat required = mass * heat capacity * temperature change

Assuming the aluminum sheet was at room temperature initially.;

Room temperature = 25 °C

Melting point of aluminum = 660.3 °C

Temperature difference = (660.3 - 25) = 635.3 903

Heat capacity of aluminum = 903 J/kg/903

Heat required = 0.05 * 903 * 635.3

Heat required = 28.7 kJ

In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.

Learn more about heat capacity at: brainly.com/question/21406849

#SPJ1

3 0
2 years ago
A star is moving away from an observer at 1% of the speed of light. At what wavelength would the observer find an emission line
Ivan

Answer:

  λ = 5940 Angstroms

Explanation:

This is an exercise of the relativistic Doppler effect

        f’= f  √((1- v / c) / (1 + v / c))

Where the speed in between the strr and the observer is positive if they move away

Let's use the relationship

         c = λ f

         f = c /λ

We replace

              c /λ’ = c /λ  √ ((1- v / c) / (1 + v / c))

              λ = λ’ √ ((1- v / c) / (1 + v / c))

Let's calculate

             v = 0.01 c

             v = 0.01 3 10⁸

             v=  3 10⁶ m / s

             λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]

             λ = 6000 √ [0.99 / 1.01]

             λ = 5940 Angstroms

6 0
3 years ago
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece
Brilliant_brown [7]

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m

or

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

8 0
2 years ago
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