16
Suppose the two masses are m1 and m2. Their initial velocities are u1 and u2 and final velocities are v1 and v2. Then, using conservation of momentum. m1*u1 + m2*u2 = m1*v1 + m2*v2 Both m1 and m2 are given. Their initial velocities u1 and u2 are given and one of the two final velocities v1 and v2 is given which leaves only one unknown. So substitute all those values and calculate away.
Answer = 2690.9776 ft21 ft = .3048 m1 ft2 = .0929 m21 m2 = 10.7639 ft2250 m2 = 2690.9776 ft2
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.
Condition of Parallelism: The Slope of two (lines) linear functions must be equal. i.e. m1=m2 Condition of perpendicularity : The product of slope of two (lines) linear functions must be equal to - 1. i.e. m1.m2=-1
16 in canadna 18 to get your m2
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?
weight, w = G * m1 * m2 / r^2, where G = the universal gravitational constant, m1 = the mass of the object, m2 = the mass of the earth, and r = the separation between the center of gravity of the object and that of the earth. w1 = G * m1 * m2 /(r1^2) -- > the weight when r = r1 w2 = G * m1 * m2 /(r2^2) -- > the weight when r = r2 Given: r2 = 4 * r1 w2 = G * m1 * m2 /(16 r1^2) = w1 / 16 Hence, the object will be 16 times lighter when it is situated 4 times farther away.
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