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3 years ago

Difference between freefall and weightlessness.

Physics

1 answer:

Margaret [11]3 years ago

6 0

Answer:

Differences between freefall and weightlessness are as follows:

<h3><u>Freefall</u></h3>

When a body falls only under the influence of gravity, it is called free fall.
Freefall is not possible in absence of gravity.
A body falling in a vacuum is an example of free fall.

<h3><u>Weightlessness</u></h3>

Weightlessness is a condition at which the apparent weight of body becomes zero.
Weightlessness is possible in absence of gravity.
A man in a free falling lift is an example of weightlessness.

Hope this helps....

Good luck on your assignment....

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A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one

Fittoniya [83]

I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)

8 0

3 years ago

Read 2 more answers

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]

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

<u>Explanation:</u>

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:

<u>For process 1:</u>

We are given:

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

Putting values in equation 1, we get:

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

<u>For process 2:</u>

We are given:

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

Putting values in equation 2, we get:

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

<u>For process 3:</u>

We are given:

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

Putting values in equation 1, we get:

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

<u>For process 4:</u>

We are given:

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

Putting values in equation 2, we get:

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

<u>For process 5:</u>

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

A woman climbs up a ladder in 1.37 s at 2.20 m/s. How tall is the ladder?

ArbitrLikvidat [17]

Answer:

The ladder is 3.014 m tall.

Explanation:

To solve this problem, we must use the following formula:

v = x/t

where v represents the woman’s velocity, x represents the distance she climbed (the height of the ladder), and t represents the time it took her to move this distance

If we plug in the values we are given for the problem, we get:

v = x/t

2.20 = x/1.37

To solve this equation for x (the height of the ladder), we must multiply both sides by 1.37. If we do this, we get:

x = (2.20 * 1.37)

x = 3.014 m

Therefore, the ladder is 3.014 m tall.

Hope this helps!

6 0

3 years ago

Read 2 more answers

PLEASE HELP DUE IN MINS. Thanks

rjkz [21]

Answer:

"C" I think....

Explanation:

I am really sorry if I am wrong, but if right, I hope this helps!

4 0

3 years ago

Read 2 more answers

Please answer all of these questions for brainly!

Nana76 [90]

Answer:

1. C
2. B
3. D
4. A

3 0

3 years ago

Difference between freefall and weightlessness.​

Difference between freefall and weightlessness.