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3 years ago

which theory of plate movements involves magma rising all the way from the lower mantle to spread apart plates

Physics

2 answers:

KatRina [158]3 years ago

6 0

Hot plumes is the answer

8_murik_8 [283]3 years ago

5 0

"
Good question!

"Which theory of plate movements involves magma rising all the way from the lower mantle to spread apart plates" is known as: "Hot Plumes" Further information may be provided at:

"A pour plate is an alternative method for using agar plates to obtain isolated colonies. Pour plates are used when it is necessary to know the number of organisms present per unit volume of specimen or other sample. When a specific aliquot is placed in the Petri dish, a count of the colonies that grow after incubation reveals their concentration in the original sample. Pour plates are used commonly in the bacteriologic examination of milk, or could also be used to determine whether sufficient bacterial numbers are present in urine samples to signify the patient has a urinary tract infection. The number of bacteria in solution can be readily quantified by using the spread plate technique. In this technique, the sample is appropriately diluted and a small aliquot transferred to an agar plate. The bacteria are then distributed evenly over the surface by a special streaking technique. After colonies are grown, they are counted and the number of bacteria in the original sample calculated. Streaking in this technique is done using a bent glass rod. 0.1 mL of bacterial suspension is placed in the center of the plate using a sterile pipet. The glass rod is sterilized by first dipping it into a 70% alcohol solution and then passing it quickly through the Bunsen burner flame. The burning alcohol sterilizes the rod at a cooler temperature than holding the rod in the burner flame thus reducing the chance of you burning your fingers. When all the alcohol has burned off and the rod has air-cooled, streak the rod back and forth across the plate working up and down several times. Unlike streaking for isolation, you want to backtrack many times in order to distribute the bacteria as evenly as possible. Turn the plate 90 degrees and repeat the side to side, up and down streaking. Turn the plate 45 degrees and streak a third time. Do not sterilize the glass rod between plate turnings. Cover the plate and wait several minutes before turning it upside down for incubation. This will allow the broth to soak into the plate so the bacteria won't drip onto the plate lid."

Quoted answers are NOT mine! I take NO credit for these answers. Answers provided at:

http://www.answers.com/Q/Which_theory_of_plate_movement_involves_magma_rising_all_the_away_from_the_...

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A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ( $v_{2,f} = 3.606\,\frac{m}{s}$ )

$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0

3 years ago

a person is sitting on the last bench can see clearly see things written on book but cannot see them distinctly on board. what t

marysya [2.9K]

Answer:

I think concave

Explanation:

7 0

3 years ago

A 0.4000 kg sample of methanol at 16.0ºC is mixed with 0.4000 kg of water at 85.0ºC. Assuming no heat loss to the surroundings,

AVprozaik [17]

Answer:

T_finalmix = 59.5 [°C].

Explanation:

In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water to methanol, at the end the temperature of the water and methanol must be equal once the thermal balance is achieved.

$Q_{water}=Q_{methanol}$

where:

$Q_{water}=m_{water}*Cp_{water}*(T_{waterinitial}-T_{final})$

mwater = mass of the water = 0.4 [kg]

Cp_water = specific heat of the water = 4180 [J/kg*°C]

T_waterinitial = initial temperature of the water = 85 [°C]

T_finalmix = final temperature of the mix [°C]

$Q_{methanol}=m_{methanol}*Cp_{methanol}*(T_{final}-T_{initialmethanol})$

Now replacing:

$0.4*4180*(85-T_{final})=0.4*2450*(T_{final}-16)\\142120-1672*T_{final}=980*T_{final}-15680\\157800=2652*T_{final}\\T_{final}=59.5[C]$

4 0

3 years ago

Question #6a) You were told to assume that the ball bounced to a height of 1.2 m and that the distance between the bounces measu

Inessa [10]

Answer:

Explanation:

In the problem they give the case of a ball that bounces on the vertical axis the height is getting smaller, on the x axis it is the same distance all the time.

Questions about the aspect of the x-axis

- Since the distance traveled between the rebounds is the same, the speed must be the same or constant throughout the entire journey.

- Since the distance and speed are equal, the time between rebounds is the same

- There can be no acceleration because the bounce gap should change, the acceleration is zero all the time

Questions about the Y axis

- The vertical speed of the first boat would be the greatest of all, so it has the highest height

b) Evidence or not Evidence

1 evidence, because the graphs are different, one is a straight line and he gives a parable

2 no evidence, nothing involved on the x axis

3 no evidence, the acceleration on the y axis is independent of the acceleration on the x axis

4 no evidence, the time that is cast is the only change that is the same for both movements

5 evidence, the fact that no mixture of components is found allows the variable to be separated into different equations

6 no evidence, says nothing about the x-axis

4 0

3 years ago

According to the Guinness Book of World Records (1990 edition, p. 169), the highest rotary speed ever attained was 2010 m/s (450

Valentin [98]

When a body performs a uniform circular motion, the direction of the velocity vector changes at every moment. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circle that gives rise to centripetal acceleration, the mathematical expression is given as,

$a = \frac{v^2}{r}$

Where,

v = Tangential Velocity

r = Radius

The linear velocity was 2010m/s in a radius of 0.159m, then the centripetal acceleration is

$a = \frac{2010^2}{0.159}$

$a = 2.54*10^7m/s^2$

Therefore the centripetal acceleration of the end of the rod is $2.54*10^7m/s^2$

7 0

3 years ago