Sedimentary rocks also known as clastic sedimentary rock
Answer:
1.75atm
Explanation:
According to Boyle's law, the pressure P of a fixed mass of gas is inversely proportional to it's volume V provided that the temperature remains constant.
![P\alpha \frac{1}{V}\\hence\\PV=constant](https://tex.z-dn.net/?f=P%5Calpha%20%5Cfrac%7B1%7D%7BV%7D%5C%5Chence%5C%5CPV%3Dconstant)
This implies the following;
Provided temperature is kept constant.
Given;
![P_1=1.5atm\\V_1=5.6L\\P_2=?\\V_2=4.8L](https://tex.z-dn.net/?f=P_1%3D1.5atm%5C%5CV_1%3D5.6L%5C%5CP_2%3D%3F%5C%5CV_2%3D4.8L)
From equation (1), we can write;
![P_1V_1=P_2V_2\\hence\\1.5*5.6=P_2*4.8\\\\P_2=\frac{1.5*5.6}{4.8}\\\\P_2=1.75atm](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2%5C%5Chence%5C%5C1.5%2A5.6%3DP_2%2A4.8%5C%5C%5C%5CP_2%3D%5Cfrac%7B1.5%2A5.6%7D%7B4.8%7D%5C%5C%5C%5CP_2%3D1.75atm)
Since all the units are consistent, there is no need for conversion.
Answer:
KE = 4 mv2 m = 2xKE valami. V m.
Explanation:
Answer:
The sound intensity of train is 1000 times greater than that of the library.
Explanation:
We have expression for sound intensity level,
![L=10log_{10}\left ( \frac{I}{I_0}\right )](https://tex.z-dn.net/?f=L%3D10log_%7B10%7D%5Cleft%20%28%20%5Cfrac%7BI%7D%7BI_0%7D%5Cright%20%29)
A train whistle has a sound intensity level of 70 dB
We have
![70=10log_{10}\left ( \frac{I_1}{I_0}\right )](https://tex.z-dn.net/?f=70%3D10log_%7B10%7D%5Cleft%20%28%20%5Cfrac%7BI_1%7D%7BI_0%7D%5Cright%20%29)
A library has a sound intensity level of about 40 dB
We also have
![40=10log_{10}\left ( \frac{I_2}{I_0}\right )](https://tex.z-dn.net/?f=40%3D10log_%7B10%7D%5Cleft%20%28%20%5Cfrac%7BI_2%7D%7BI_0%7D%5Cright%20%29)
Dividing both equations
![\frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000](https://tex.z-dn.net/?f=%5Cfrac%7B70%7D%7B40%7D%3D%5Cfrac%7B10log_%7B10%7D%5Cleft%20%28%20%5Cfrac%7BI_1%7D%7BI_0%7D%5Cright%20%29%7D%7B10log_%7B10%7D%5Cleft%20%28%20%5Cfrac%7BI_2%7D%7BI_0%7D%5Cright%20%29%7D%5C%5C%5C%5C%5Cfrac%7B7%7D%7B4%7D%3D%5Cfrac%7Blog_%7B10%7D%5Cleft%20%28%20%5Cfrac%7BI_1%7D%7BI_0%7D%5Cright%20%29%7D%7Blog_%7B10%7D%5Cleft%20%28%20%5Cfrac%7BI_2%7D%7BI_0%7D%5Cright%20%29%7D%5C%5C%5C%5C10%5E7%5Cfrac%7BI_2%7D%7BI_0%7D%3D10%5E4%5Cfrac%7BI_1%7D%7BI_0%7D%5C%5C%5C%5C%5Cfrac%7BI_1%7D%7BI_2%7D%3D10%5E3%3D1000)
The sound intensity of train is 1000 times greater than that of the library.
For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.