Answer:


Explanation:
m = Mass of proton = 
v = Speed of proton = 0.5c = 
Circumference of the colider is 7 km


Centripetal acceleration is 

Force on protons is 
We can calculate this with the law of conservation of energy. Here we have a food package with a mass m=40 kg, that is in the height h=500 m and all of it's energy is potential. When it is dropped, it's potential energy gets converted into kinetic energy. So we can say that its kinetic and potential energy are equal, because we are neglecting air resistance:
Ek=Ep, where Ek=(1/2)*m*v² and Ep=m*g*h, where m is the mass of the body, g=9.81 m/s² and h is the height of the body.
(1/2)*m*v²=m*g*h, masses cancel out and we get:
(1/2)*v²=g*h, and we multiply by 2 both sides of the equation
v²=2*g*h, and we take the square root to get v:
v=√(2*g*h)
v=99.04 m/s
So the package is moving with the speed of v= 99.04 m/s when it hits the ground.
50% of the moon is always illuminated, however during it's quarter phase means that we only see a quarter of what's really lit up. So it LOOKS like the moon is only 25% lit and 75% dark, it's truly 50/50. We only see that 25% since we can see it from one angle.
Answer: 24.97 kg
Explanation:
The gravitational force between two objects of masses M1, and M2 respectively, and separated by a distance R, is:
F = G*(M1*M2)/R^2
Where G is the gravitational constant:
G = 6.67*10^-11 m^3/(kg*s^2)
In this case, we know that
R = 0.002m
F = 0.0104 N
and that M1 = M2 = M
And we want to find the value of M, then we can replace those values in the equation to get
0.0104 N = (6.67*10^-11 m^3/(kg*s^2))*(M*M)/(0.002m)^2
(0.0104 N)*(0.002m)^2/(6.67*10^-11 m^3/(kg*s^2)) = M^2
623.69 kg^2 = M^2
√(623.69 kg^2) = M = 24.97 kg
This means that the mass of each object is 24.97 kg
Answer:
Angle: 
Explanation:
<u>Two-Dimension Motion</u>
When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.
Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.
To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

where
is the speed of the boy in still water and
is the angle respect to the shoreline. If the river flows at speed
, we now set


