Explanation:
One of the takeaways of the General Theory of Relativity (GTR) was that the light can be curved because of the gravity of a massive object. Einstein had proposed the idea of <em>space-time</em> fabric. Every object having mass will create depression in this fabric. Heavier the object, bigger the depression. Thus when light will pass near a heavy object lets say our Sun, it will deflect. He also gave mathematical formula to calculate the deflection.
The same was proved during the Total Solar Eclipse of 29 May 1919. Two scientists named Arthur Eddington and Frank Dyson conducted an experiment. In this eclipse the Sun was to be in front of Hyades in Taurus constellation. They took the measurement of stars of Hyades visible during the eclipse and then compared them with the actual readings. The deflection was clearly visible and the amount of deflection was very close to the values predicted by General Theory of Relativity. Thus they proved the theory right.
Answer:
6 m/s²
Explanation:
We are given;
- The mass of the block as 6.2 kg
- Horizontal force as 50.9 N
- Frictional force as 13.7 N
We are required to determine the acceleration of the block.
- According to second Newton's Law of motion; Resultant force is directly proportional to the rate of change in linear momentum.
Therefore;
F = Ma , where F is the resultant force, m is the mass and a is the acceleration.
In this case;
Resultant force, F = Horizontal force - friction force
= 50.9 N - 13.7 N
= 37.2 N
Thus;
37.2 N = 6.2 kg × a
Hence;
acceleration, a = 37.2 N ÷ 6.2 kg
= 6 m/s²
Therefore, the acceleration of the block is 6 m/s²
There are supposed to be numbers in these places:
" _____ of methane"
" ... pressure of exactly _____ "
" ... temperature of _____ "
I've seen a lot of questions with missing numbers before, but I think this is the first time I ever saw a question where even the number of significant digits to round the answer to is missing.
"Round your answer to _____ significant digits."
Answer:
U = 5.37*10^33 J
Explanation:
The gravitational potential energy between two bodies is given by:
![U_{1,2}=-G\frac{m_1m_2}{r_{1,2}}](https://tex.z-dn.net/?f=U_%7B1%2C2%7D%3D-G%5Cfrac%7Bm_1m_2%7D%7Br_%7B1%2C2%7D%7D)
G: Cavendish's constant = 6.67*10^-11 m^3/kg.s
For three bodies the total gravitational potential energy is:
![U_{T}=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_{T}=-G[\frac{m_1m_2}{r_{1,2}}+\frac{m_1m_3}{r_{1,3}}+\frac{m_2m_3}{r_{2,3}}]](https://tex.z-dn.net/?f=U_%7BT%7D%3DU_%7B1%2C2%7D%2BU_%7B1%2C3%7D%2BU_%7B2%2C3%7D%5C%5C%5C%5CU_%7BT%7D%3D-G%5B%5Cfrac%7Bm_1m_2%7D%7Br_%7B1%2C2%7D%7D%2B%5Cfrac%7Bm_1m_3%7D%7Br_%7B1%2C3%7D%7D%2B%5Cfrac%7Bm_2m_3%7D%7Br_%7B2%2C3%7D%7D%5D)
BY replacing the values of the parameters for 1->earth, 2->moon and 3->sun you obtain:
![U_{T}=-(6.67*10^{-11}m^3/kg.s)[\frac{(5.98*10^{24}kg)(7.36*10^{22}kg)}{3.84*10^{8}m}+\\\\\frac{(5.98*10^{24}kg)(1.99*10^{30}kg)}{1.496*10^{11}m}+\frac{(7.36*10^{22}kg)(1.99*10^{30}kg)}{1.496*10^{11}m-3.84*10^8m}]\\\\U_{T}=5.37*10^{33}J](https://tex.z-dn.net/?f=U_%7BT%7D%3D-%286.67%2A10%5E%7B-11%7Dm%5E3%2Fkg.s%29%5B%5Cfrac%7B%285.98%2A10%5E%7B24%7Dkg%29%287.36%2A10%5E%7B22%7Dkg%29%7D%7B3.84%2A10%5E%7B8%7Dm%7D%2B%5C%5C%5C%5C%5Cfrac%7B%285.98%2A10%5E%7B24%7Dkg%29%281.99%2A10%5E%7B30%7Dkg%29%7D%7B1.496%2A10%5E%7B11%7Dm%7D%2B%5Cfrac%7B%287.36%2A10%5E%7B22%7Dkg%29%281.99%2A10%5E%7B30%7Dkg%29%7D%7B1.496%2A10%5E%7B11%7Dm-3.84%2A10%5E8m%7D%5D%5C%5C%5C%5CU_%7BT%7D%3D5.37%2A10%5E%7B33%7DJ)
hence, the total gravitational energy is 5.37*10^33 J
Answer: demesne lands
Explanation:
" (demesne lands) which were for the personal use of the lord of the manor."