A 0.7 kg object is at rest. A 6.5 N force to the right acts on the object during a time interval of 3.7 seconds. What is the velocity of the object at the end of this interval?

F = ma

6.5 = .7a
a = 6.5/.7

v = at = 6.5/.7 * 3.7 = 34.357

To find the velocity of the object at the end of the interval, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration:

F = m * a

In this case, the mass of the object is given as 0.7 kg, and the force is given as 6.5 N. We can rearrange the formula to solve for acceleration:

a = F / m

Substituting the given values, we get:

a = 6.5 N / 0.7 kg
a ≈ 9.29 m/s^2

Now, we can use the equation of motion to find the velocity. The equation is:

v = u + a * t

Where:
v = final velocity
u = initial velocity (which is 0 in this case, as the object is at rest)
a = acceleration
t = time interval

Substituting the values, we get:

v = 0 + 9.29 m/s^2 * 3.7 s
v ≈ 34.33 m/s

Therefore, the velocity of the object at the end of the 3.7-second interval is approximately 34.33 m/s to the right.