An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 units/sec. what is the force on the object due to its acceleration at left parenthesis 2 superscript 1 divided by 2 baseline comma 2 right parenthesis? (remember newton's law, fequalsma.)

Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2).  Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature.  This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~