A1190 kg car pulls a 460 kg trailer. the car exerts a horizontal force of 4070 n against the ground in order to accelerate. what force does the car exert on the trailer? assume an effective friction coefficient of 0.190 for the trailer.

F= 4070 N ( the force of the car )
m = 1190 kg
M = 460 kg 
k ( or usually the Greek letter:"mi")=0.190 
T = ? ( the force on the trailer )
F = ( M +m ) a + k m g
T = m a + k m g
a = ?
a = (F - k m g)/ (M + m )
T = m · (F - k m g ) / ( M + m ) + k m g
T = 1190 kg·( 4070 N-0.190·1190 kg·9.81 m/s²):(460kg+1190kg)+
+0190·1190kg·9.81m/s²=
= 1190 kg ·1851.96 N :1650 kg + 2218.04 N = 
=1335.66 N + 2218.04 N = 3553.7 N
The force that car exerts on the trailer is 3553.7 N.

Answer: buoyancy forceexplanation: in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.