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3 years ago

Was Wegener able to provide a mechanism as to how the plates moved accordingly? Why or why not?

Physics

1 answer:

Marina86 [1]3 years ago

3 0

Answer:

The main issue with Wegener's Continental Drift Theory was he did not have a mechanism behind the drifting of continents. He had a substantial amount of evidence that made sense; nonetheless, without a driving force behind it, the scientific community simply discredited his entire idea.

Explanation:

You might be interested in

As a raindrop falls from a cloud to the surface of Earth,

Sunny_sXe [5.5K]

The correct answer would be C. it moves at a constant speed. The troposphere(the layer our weather is in) is not nearly high enough for gravity to be different at different altitudes.

6 0

3 years ago

Acar accelerates from 4 meters/second to 16 meters/second in 4 seconds. The car's acceleration is

s2008m [1.1K]

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: $\frac{∆v}{t}$

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 $\frac{feet}{second}$ then sped up to 5 $\frac{feet}{second}$ , then the change in my speed would be 2 $\frac{feet}{second}$ , because I started walking 2 $\frac{feet}{second}$ faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 $\frac{meters}{second}$ to 16 $\frac{meters}{second}$ in 4 seconds. What would the acceleration be? Let's set up an equation:

a = $\frac{∆v}{t}$

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 $\frac{meters}{second}$ by 4 $\frac{meters}{second}$ . That makes sense, right? Back to the equation.

a = $\frac{∆v}{t}$
a = $\frac{16-4}{4}$

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = $\frac{12}{4}$

a = 3 ( $\frac{meters}{second^{2}}$ )

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit $\frac{\frac{meters}{second}}{second}$ , which can be simplified down to $\frac{meters}{second^{2} }$ )

Let me know if you need clarification on anything I explained here!
- breezyツ

6 0

2 years ago

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

Answer: The change in entropy of the given process is 1324.8 J/K

Explanation:

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

$\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})$ .......(1)

where,

$\Delta S$ = Entropy change

$C_{p,m}$ = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)

$T_2$ = final temperature

$T_1$ = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

where,

$\Delta S$ = Entropy change

m = mass of ice

$\Delta H_{f,v}$ = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

For process 1:

We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

For process 2:

We are given:

$m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K$

For process 3:

We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

$\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K$

For process 4:

We are given:

$m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K$

For process 5:

We are given:

$m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K$

Putting values in equation 1, we get:

$\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K$

Total entropy change for the process = $\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5$

Total entropy change for the process = $[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K$

Hence, the change in entropy of the given process is 1324.8 J/K

4 0

3 years ago

Take a look at the weather map. The front seen there causes short periods of storms and heavy rains. What type of front is this?

erastova [34]

I am pretty sure it is a cold front.

5 0

3 years ago

Read 2 more answers

Who much force is needed to accelerate a 68 kilogram-skier at a Rate of 1.2 m/sec

liberstina [14]

If F =m*a
and the question says how much force the s needed to accelerate a 68kg skier to a rate of 1.2ms^-2
Then F = 68*1.2

7 0

3 years ago