A 5.57-kg object passes through the origin at time t = 0 such that its x component of velocity is 5.25 m/s and its y component of velocity is -3.15 m/s.

(a) What is the kinetic energy of the object at this time?

(b) At a later time t = 2.00 s, the particle is located at x = 8.50 m and y = 5.00 m. What constant force acted on the object during this time interval?
magnitude

magnitude
Degrees?

(c) What is the speed of the particle at t = 2.00 s?
m/s

A) i got correct which is 104

B) i got 10.5
and 30.5 degrees which its telling me im off by10%

C) i need be corrected i think

C) To find the speed of the particle at t = 2.00 s, we can use the formula for speed, which is the magnitude of the velocity vector.

The velocity vector has two components: an x-component and a y-component. We are given the x and y coordinates of the particle at t = 2.00 s, so we can calculate the displacement in the x and y directions.

Using the formula for displacement in the x-direction: Δx = x - x₀ = 8.50 m - 0 m = 8.50 m
Using the formula for displacement in the y-direction: Δy = y - y₀ = 5.00 m - 0 m = 5.00 m

Now, we can calculate the magnitude of the displacement vector using the Pythagorean theorem:
|Δr| = sqrt(Δx² + Δy²)
|Δr| = sqrt((8.50 m)² + (5.00 m)²)
|Δr| = sqrt(72.25 m² + 25.00 m²)
|Δr| = sqrt(97.25 m²)
|Δr| = 9.86 m

Since the particle's velocity vector is equivalent to its displacement vector divided by the time interval, we can calculate the speed as:
speed = |Δr| / Δt
speed = 9.86 m/ 2.00 s
speed = 4.93 m/s

Therefore, at t = 2.00 s, the speed of the particle is 4.93 m/s.