Answer:Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.
Explanation:
Below is the answer. I hope it help.
T ( t ) = C e k t + T m where Tm is the temperature of the surroundings
T ( t ) = C e k t + T m
T ( 0 ) = 20
T ( 1 ) = T ( 0 ) + 2 = 22
C + T m = 20 C+Tm=20
C e k + T m = 22
Comets are like "dirty snowballs"; frozen gasses with dust and rocks in them. Each pass near the Sun causes the comet's nucleus to be exposed to intense sunlight, which causes some tiny fraction of the gas to evaporate and carry some of the dust and rock away into space. The gas and dust, near the Sun, cause the comet's "tail", and repeated passes cause dust and rock to spread out along most of the orbit of a comet. When the Earth enters one of these trails of old comet dust, we have meteor showers.
<span>On rare occasions, comets break apart or even more rarely, crash into planets. In 1994, the comet Shoemaker-Levy 9 broke apart and then collided with the planet Jupiter.</span>
Answer:
D
Explanation:
Cá xương, chim, thú, cá sấu không có sự pha trộn máu giàu O2 và máu giàu CO2 ở tim vì tim cá có 2 ngăn, tim các loài chim, thú, cá sấu có 4 ngăn
Answer:
Explanation:
a. To find the gravitational potential energy you take into account the potential energy generated by a piece of mass in the disk. You also take into account that the gravitational field at point A points to the center of the disk. Then, you can consider an element of the force, Fx:
m: mass of a body at a distance of "a" perpendicular to the disk.
R:radius of the disk
M: mass of the disk
G: Cavendish's constant
by solving the integral you obtain:
![F_x=2\frac{GmM}{R^2}[1-cos\theta]](https://tex.z-dn.net/?f=F_x%3D2%5Cfrac%7BGmM%7D%7BR%5E2%7D%5B1-cos%5Ctheta%5D)
![F_x=2GmM[1-\frac{a}{\sqrt{R^2+a^2}}]](https://tex.z-dn.net/?f=F_x%3D2GmM%5B1-%5Cfrac%7Ba%7D%7B%5Csqrt%7BR%5E2%2Ba%5E2%7D%7D%5D)
(1)
To find the gravitational energy you use:
![U=-\int F_x dx=-\int[2GmM(1-\frac{x}{\sqrt{R^2+x^2}})]dx\\\\U=-2GmM[x+\sqrt{R^2+x^2}]\\\\U=-2GmM[a+\sqrt{R^2+a^2}]](https://tex.z-dn.net/?f=U%3D-%5Cint%20F_x%20dx%3D-%5Cint%5B2GmM%281-%5Cfrac%7Bx%7D%7B%5Csqrt%7BR%5E2%2Bx%5E2%7D%7D%29%5Ddx%5C%5C%5C%5CU%3D-2GmM%5Bx%2B%5Csqrt%7BR%5E2%2Bx%5E2%7D%5D%5C%5C%5C%5CU%3D-2GmM%5Ba%2B%5Csqrt%7BR%5E2%2Ba%5E2%7D%5D)
you replace the values of the parameters in the point A:
![U=-2(6.67*10^{-11})(250)(5.98*10^{24})[(7.83*10^6)+(\sqrt{(7.83*10^6)^2+(6.25*10^7)^2})]\\\\U=-1.41*10^{25}J](https://tex.z-dn.net/?f=U%3D-2%286.67%2A10%5E%7B-11%7D%29%28250%29%285.98%2A10%5E%7B24%7D%29%5B%287.83%2A10%5E6%29%2B%28%5Csqrt%7B%287.83%2A10%5E6%29%5E2%2B%286.25%2A10%5E7%29%5E2%7D%29%5D%5C%5C%5C%5CU%3D-1.41%2A10%5E%7B25%7DJ)
b. For point B you have a=0.
![U=-2(6.67*10^{-11})(250)(5.98*10^{24})[(7.83*10^6)+(\sqrt{((7.83*10^6)^2})]\\\\U=-3.12*10^{24}J](https://tex.z-dn.net/?f=U%3D-2%286.67%2A10%5E%7B-11%7D%29%28250%29%285.98%2A10%5E%7B24%7D%29%5B%287.83%2A10%5E6%29%2B%28%5Csqrt%7B%28%287.83%2A10%5E6%29%5E2%7D%29%5D%5C%5C%5C%5CU%3D-3.12%2A10%5E%7B24%7DJ)
c. To find the kinetic energy you use:
However, the net work is also the difference between the gravitaional potential energies calculated in the previous steps.
