The surface is frictionless, Two blocks in contact with each other move

to the right across a horizontal surface by the

two forces shown

Determine the magnitude of the force exerted on the block with mass 1.4 kg by the

block with mass 6.4 kg.

Answer in units of N

9514 1404 393

  55 8/39 N ≈ 55 N

Step-by-step explanation:

The net force on the 1.4 kg block is proportional to its mass. That is, the ratio of net force on that block to the net force on the total mass is the same as the ratio of the block's mass to the total mass.

  net force on 1.4 kg = (1.4 kg)/(1.4 +6.4 kg) × (64 N -15 N) = (7/39)(49 N)

  net force on 1.4 kg = 8 31/39 N

Then the force applied by the 6.4 kg mass is ...

  64 N - 8 31/39 N = 55 8/39 N ≈ 55 N . . . . applied by 6.4 kg mass

it reflected because of the glass maube

step-by-step explanation:

step-by-step explanation:

you can add the two numbers and divide the result by 2. or you can find the gap between the two numbers and add half of this to the smaller number.