In Kepler's laws of planetary motion, m1 and m2 represent the masses of two objects (usually the Sun and a planet) that are orbiting around each other. Kepler's laws describe the relationship between the orbit of a planet and the mass of the objects involved.
The force exerted on m2 due to the gravitational attraction with m1 is equal in magnitude but opposite in direction to the force exerted on m1. This is known as Newton's third law of motion - for every action, there is an equal and opposite reaction.
The attraction between two objects is: G x m1 x m2 / distance squared where m1 and m2 are the two masses involved, and G is the gravitational constant - approximately 6.674 x 10 to the power -11, in SI units.
M2 includes M1 components (currency, demand deposits) along with savings accounts, time deposits, and non-institutional money market funds.
M2 and M1 are measures of the money supply. M1 includes physical money, such as paper currency and coins, as well as demand deposits and other liquid assets that can be quickly converted into cash. M2 includes all of M1 plus savings deposits, time deposits, and other non-cash assets that can be easily converted into cash.
The velocities of the two bodies after the elastic collisions are given by V1=(M1-M2)U1/(M1+M2)+2M2U2/(M1+M2) V2=(M2-M1)U2/(M1+M2)+2U1M1/(M1+M2) Where, V1,V2 are the velocities of the two bodies after collision. U1,U2 are the velocities of the two bodies before colision.(U1>U2) M1,M2 are the masses of the two bodies. when the mass of two bodies are equal that is M1= M2 then V1=0+2MU2/2M=U2 V2=0+2MU1/2M=U1 Thus when two billiard balls of equal masses undergo perfectly elastic collision the velocities the two bodies are interchanged after the collision.
because 3>2>1 ? Other than that, depends on what m1,m2 and m3 represent.
What is?? F = G m1 m2 / d2
What is?? F = G m1 m2 / d2
What is the difference between M1 and M2?
If the slopes are m1 and m2 then m1*m2 = -1 or m2 = -1/m1.
if(m1>m2) f=m1; s=(m2>m3)?m1!m3 what its meaning of this?
Law of Gravity: Fg = G(m1*m2)/r^2
The force exerted on m2 due to the gravitational attraction with m1 is equal in magnitude but opposite in direction to the force exerted on m1. This is known as Newton's third law of motion - for every action, there is an equal and opposite reaction.
F = G M1M2/R2F and R are vectors.G, M1, M2 are scalars.
Let us suppose we are plotting y vs x and obtain a straight line. Then we pick a set of two coordinates, x1,y1 and x2,y2 The slope, M, is then given by the equation M (y2-y1)/(x2-x1) If we apply this to a force vs mass graph, we obtain the expression M (F2-F1)/(m2-m1),but F ma according to Newton's second law, where a is the acceleration, which leads to (m2a2-m1a1)/(m2-m1), but if a2 a1 a, as it will if the line is straight, then M a(m2-m1)/(m2-m1) a, so the slope, M, of your graph is acceleration.
The force, written as an equation, is:F = G (m1)(m2) / r2, whereF is the Force between the massesG is the gravitational constant (~= 6.674 x 10-11 N m2/kg2)m1 is one of the massesm2 is the other massr is the distance between the masses (center to center)Take the formula, and solve for r (I'll show the steps): Fold = G (m1)(m2) / r2.(r2)(Fold)= G (m1)(m2)(r2)= G (m1)(m2) / (Fold)r= √ [ G (m1)(m2) / (Fold) ]Plug the formula into itself, but remember, r = 3r (it tripled).Fnew= G (m1)(m2) / (3r)2.Fnew= G (m1)(m2) /(3√ [ G (m1)(m2) / (Fold) ])2.Fnew=G (m1)(m2)/(32G (m1)(m2) / (Fold) )
this procedure work for ternary search int tsearch(int *a,int i,int j,int k) { int m1,m2,len; len = j - i + 1 ; m1=i + (int)floor((float)(len))/3; m2=i + (int)ceil((float)(len))/3; if(k==a[m1]) { printf("\nno found at %d",m1); return m1; } else if(k==a[m2]) { printf("\nno found at %d",m2); return m2; } if(len!= 0) { if(k<a[m1]) return(tsearch(a,i,m1-1,k)); if(k>a[m2]) return(tsearch(a,m2+1,j,k)); } else return -1 ; }