Calculate the tension in a horizontal string that whirls a 1.9-kg toy in a circle of radius 2.3m when it moves at 2.8m/s on an icy surface.

Express your answer to two significant figures and include the appropriate units

m v^2/r = 1.9 (2.8)^2/2.3 Newtons

12.30 N

To calculate the tension in the horizontal string, we can start by considering the forces acting on the toy. The tension in the string is the force that keeps the toy moving in a circular path. This force must provide the centripetal force required for circular motion.

The centripetal force is given by the equation:

F = (m * v^2) / r

Where:
F is the centripetal force,
m is the mass of the toy,
v is the velocity of the toy, and
r is the radius of the circular path.

In this case, the mass of the toy (m) is 1.9 kg, the velocity (v) is 2.8 m/s, and the radius (r) is 2.3 m.

Let's substitute these values into the equation:

F = (1.9 kg * (2.8 m/s)^2) / 2.3 m

Calculating this expression, we have:

F = (1.9 * 2.8^2) / 2.3

F = (1.9 * 7.84) / 2.3

F ≈ 6.616 / 2.3

F ≈ 2.872

Rounding this value to two significant figures, the tension in the horizontal string is approximately:

T = 2.9 N

Therefore, the tension in the horizontal string is 2.9 N (Newtons).