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.
The force of gravity is F=G*m1*m2/r^2 G is the universal gravitation constant 6.67*10^-11 m^3kg^-1s^-2 m1, m2 are the masses of the two objects, r is the separation. The force on m1 acts in the direction of m2, and the force on m2 acts in the direction of m1.
Alright, so basically you're given: m1+m2=4.0kg, r=25m, Fg=2.5*10-10N We know that Fg=Gm1m2/r2 Using the fact that m1+m2=4.0kg, we know that m2=4.0kg-m1 So: Fg=Gm1(4.0kg-m1)/r2 =G(m1(4.0)-m12)/r2 So: (-G/r2)m12+(4G/r2)m1-Fg=0 Here we can use the quadratic formula to solve for m1 Once you have m1, just use m1+m2=4.0kg to find m2 I'll leave the actual number crunching to you. Hope this helps!
if we assume m1 (mass) and v1 (velocity) for first mass , m2 and v2 for second mass ,we have : m1 v1i + m2 v2i = m1 v1f + m2 v2f and 1/2 m1 v1i2 + 1/2 m2 v2i2 = 1/2 m1 v1f2 +1/2 m2 v2f2 i : initial f : final This is a simplified version: vf1= ((m1-m2)/(m1+m2))(v1i)+ (2m2/(mi+m2)vi2
m1 + m2 = 5 kg F = 1 x 10-8 N G = 6.67 x 10-11 N m2/kg2 r = 0.2 m F = (Gm1m2) / r2 1 x 10-8 N = (6.67 x 10-11 N m2/kg2)(m1)(5-m1) divide by 6.67 x 10-11 N m2/kg2 6 = 5m1 - m12 m12 - 5m1 + 6 = 0 (m1 - 3)(m1- 2) = 0 m1 = 3 kg m2 = 2 kg
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?
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 ; }
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.
The force of gravity is F=G*m1*m2/r^2 G is the universal gravitation constant 6.67*10^-11 m^3kg^-1s^-2 m1, m2 are the masses of the two objects, r is the separation. The force on m1 acts in the direction of m2, and the force on m2 acts in the direction of m1.
Cash is part of M1.
permotion list of m2 officers
M 1
M1 is what is outside the banking system: Your cash, coins, your checking account. M2 is: All of M1 plus, savings accounts, money in banks, small time deposits...etc .