A net force of 25N is applied for 5.7s to a 12-kg box initially at rest. Waht is the speed of the box at the end of the 5.7s interval?

a = Fn/m = 25 / 12 = 2.08 m/s^2.

Va*t = 2.08m/s^2 * 5.7s. = 11.88 m/s.

12 m/s

A 10-N net force is applied for 5.0 s to a 12-kg box initially at rest. What is the speed of the box at the end of the 5.0-s interval?

Well, let's calculate that, shall we? We know that Force (F) equals mass (m) times acceleration (a). In this case, the mass is 12 kg and the force is 25 N, so we can rearrange the equation to find the acceleration:

a = F / m

a = 25 N / (12 kg)

a ≈ 2.08 m/s²

Now, we can use a different equation to find the final velocity (v) of the box:

v = u + at

where u is the initial velocity, which is 0 m/s because the box was initially at rest, and t is the time, which is 5.7s:

v = 0 m/s + (2.08 m/s²)(5.7 s)

v ≈ 11.856 m/s

So, after 5.7 seconds, the speed of the box will be approximately 11.856 m/s. That's moving faster than I can juggle pies!

To find the speed of the box at the end of the 5.7s interval, we can use the formula for calculating speed:

Speed = (Force * Time) / Mass

Step 1: Identify the given values:
Force = 25 N (net force applied)
Time = 5.7 s (duration of force application)
Mass = 12 kg (mass of the box)

Step 2: Plug the values into the formula:
Speed = (25 N * 5.7 s) / 12 kg

Step 3: Calculate the result:
Speed = (142.5 N·s) / 12 kg
Speed = 11.875 m/s

Therefore, the speed of the box at the end of the 5.7s interval is 11.875 m/s.

that is correct