Answer: by definition work = Force * distanceForce = 100Ndistance = 6 - 2 = 4work = 100 * 4 = 400 J or CStill stuck? Get 1-on-1 help from an expert tutor now.
Explanation:
Explanation:
Given parameters:
Mass of Neil Armstrong = 160kg
Gravitational pull of earth = 10N/kg
Moon's pull = 17% of the earth's pull
Unknown:
Difference between Armstrong's weight on moon and on earth.
Solution:
To find the weight,
Weight = mass x acceleration due to gravity = mg
Moon's gravitational pull = 17% of the earth's pull = 17% x 10 = 1.7N/kg
Weight on moon = 160 x 1.7 = 272N
Weight on earth = 160 x 10 = 1600N
The difference in weight = 1600 - 272 = 1328N
The weight of Armstrong on earth is 1328N more than on the moon.
Learn more:
Weight and mass brainly.com/question/5956881
#learnwithBrainly
For the answer to the question above,
the distance from i to j is 5 parts
(2 parts from i to k and 3 parts from k to j)
The y distance from i to j is
10 - 2 = 8
Each part is 8/5 = 1.6
Therefore the distance between the 2 parts from i to k is 3.2
From the y coordinate of I which is 2 plus the 3.2 to point k
2 + 3.2 = 5.2
Answer y =5.2
Now just convert that to fraction and that will be the answer
Answer:0.061
Explanation:
Given
![T_C=300 k](https://tex.z-dn.net/?f=T_C%3D300%20k)
Temperature of soup ![T_H=340 K](https://tex.z-dn.net/?f=T_H%3D340%20K)
heat capacity of soup ![c_v=33 J/K](https://tex.z-dn.net/?f=c_v%3D33%20J%2FK)
Here Temperature of soup is constantly decreasing
suppose T is the temperature of soup at any instant
efficiency is given by
![\eta =\frac{dW}{Q}=1-\frac{T_C}{T}](https://tex.z-dn.net/?f=%5Ceta%20%3D%5Cfrac%7BdW%7D%7BQ%7D%3D1-%5Cfrac%7BT_C%7D%7BT%7D)
![dW=Q(1-\frac{T_C}{T})](https://tex.z-dn.net/?f=dW%3DQ%281-%5Cfrac%7BT_C%7D%7BT%7D%29)
![dW=c_v(1-\frac{T_C}{T})dT](https://tex.z-dn.net/?f=dW%3Dc_v%281-%5Cfrac%7BT_C%7D%7BT%7D%29dT)
integrating From
to ![T_C](https://tex.z-dn.net/?f=T_C)
![\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT](https://tex.z-dn.net/?f=%5Cint%20dW%3D%5Cint_%7BT_C%7D%5E%7BT_H%7Dc_v%281-%5Cfrac%7BT_C%7D%7BT%7D%29dT)
![W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT](https://tex.z-dn.net/?f=W%3D%5Cint_%7BT_C%7D%5E%7BT_H%7D33%5Ccdot%20%281-%5Cfrac%7B300%7D%7BT%7D%29dT)
![W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20T-T_C%5Cln%20T%5Cright%20%5D_%7BT_H%7D%5E%7BT_C%7D)
![W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D)
Now heat lost by soup is given by
![Q=c_v(T_C-T_H)](https://tex.z-dn.net/?f=Q%3Dc_v%28T_C-T_H%29)
Fraction of the total heat that is lost by the soup can be turned is given by
![=\frac{W}{Q}](https://tex.z-dn.net/?f=%3D%5Cfrac%7BW%7D%7BQ%7D)
![=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D%7D%7Bc_v%28T_C-T_H%29%7D)
![=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}](https://tex.z-dn.net/?f=%3D%5Cfrac%7BT_C-T_H-T_C%5Cln%20%28%5Cfrac%7BT_C%7D%7BT_H%7D%29%7D%7BT_C-T_H%7D)
![=\frac{300-340-300\ln (\frac{300}{340})}{300-340}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B300-340-300%5Cln%20%28%5Cfrac%7B300%7D%7B340%7D%29%7D%7B300-340%7D)
![=\frac{-40+37.548}{-40}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B-40%2B37.548%7D%7B-40%7D)
![=0.061](https://tex.z-dn.net/?f=%3D0.061)
The correct answers are as follows:
<span>1) hydrogenous sediment
2)sand and gravel
3) They rapidly break down at surface temperatures and pressures.</span>