Answer:
1.63 N
Explanation:
F = GMm/r^2
= (6.67x10^-11)(10x10^5)(3x10^5) / 3.5^2
= 1.63 N ( 3 sig. fig.)
Answer:
There is a loss of fluid in the container of 0.475L
Explanation:
To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.
The formula that describes this thermal expansion process is given by:
![\Delta V = \beta V_0 \Delta T](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%5Cbeta%20V_0%20%5CDelta%20T)
Where,
Change in volume
Initial Volume
Change in temperature
coefficient of volume expansion (Coefficient of copper and of the liquid for this case)
There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,
![\Delta V_T = \Delta V_l - \Delta V_c](https://tex.z-dn.net/?f=%5CDelta%20V_T%20%3D%20%5CDelta%20V_l%20-%20%5CDelta%20V_c)
Where,
= Change in the volume of liquid
= Change in the volume of copper
Then replacing with the previous equation we have:
![\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%5Cbeta_l%20V_0%20%5CDelta%20T-%20%5Cbeta_c%20V_0%20%5CDelta%20T)
![\Delta V = (\beta_l-\beta_c)V_0\Delta T](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%28%5Cbeta_l-%5Cbeta_c%29V_0%5CDelta%20T)
Our values are given as,
Thermal expansion coefficient for copper and the liquid to 20°C is
![\beta_c = 51*10^{-6}/\°C](https://tex.z-dn.net/?f=%5Cbeta_c%20%3D%2051%2A10%5E%7B-6%7D%2F%5C%C2%B0C)
![\beta_l = 400*10^{-6}/\°C](https://tex.z-dn.net/?f=%5Cbeta_l%20%3D%20400%2A10%5E%7B-6%7D%2F%5C%C2%B0C)
![V_0 = 16L](https://tex.z-dn.net/?f=V_0%20%3D%2016L)
![\Delta T = (95\°C-10\°C)](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%20%2895%5C%C2%B0C-10%5C%C2%B0C%29)
Replacing we have that,
![\Delta V = (\beta_l-\beta_c)V_0\Delta T](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%28%5Cbeta_l-%5Cbeta_c%29V_0%5CDelta%20T)
![\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%28400%2A10%5E%7B-6%7D%2F%5C%C2%B0C-51%2A10%5E%7B-6%7D%2F%5C%C2%B0C%29%2816L%29%2895%5C%C2%B0C-10%5C%C2%B0C%29)
![\Delta V = 0.475L](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%200.475L)
Therefore there is a loss of fluid in the container of 0.475L
Answer:
Reading a Graduated Cylinder
Place the graduated cylinder on a flat surface and view the height of the liquid in the cylinder with your eyes directly level with the liquid. The liquid will tend to curve downward. This curve is called the meniscus. Always read the measurement at the bottom of the meniscus.....
hope it helps....
Walk out. If it's denser than air, it'll settle to the bottom
( 1.05 x 10¹⁵ km ) x ( 1 LY / 9.5 x 10¹² km ) x ( 1 psc / 3.262 LY ) =
(1.05) / (9.5 x 3.262) x (km · LY · psc) / (km · LY) x (10¹⁵⁻¹²) =
(0.03388) x (psc) x (10³) =
33.88 parsecs