A 2.85 kg stationary package explodes into three parts that then slide across a frictionless floor. The package had been at the origin of a coordinate system. Part 1 has mass m1 = 0.500 kg and velocity (10.0 + 12.0) m/s. Part 2 has mass m2 = 0.750 kg, a speed of 13.5 m/s, and travels at an angle 115° (counterclockwise from the positive direction of the x axis.)

(a) What is the speed of part 3?
_________m/s

(b) In what direction does it travel?
___________° counterclockwise from the positive direction of the x axis

An elevator cable can withstand a maximum tension of 1.89×104N before breaking. The elevator has a mass of 480kg and a maximum acceleration of 2.25m/s^2 . Engineering safety standards require that the cable tension never exceed two-thirds of the breaking tension.

To solve this problem, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before the explosion is equal to the total momentum after the explosion.

Let's start by calculating the x- and y-components of the total momentum before the explosion.

Before the explosion, the package is stationary, so its momentum is zero in both the x and y directions.

In the x-direction:
Total momentum before explosion = m1 * vx1 + m2 * vx2

In the y-direction:
Total momentum before explosion = m1 * vy1 + m2 * vy2

Given values:
m1 = 0.500 kg
vx1 = 10.0 m/s
vy1 = 12.0 m/s
m2 = 0.750 kg
v2 = 13.5 m/s
θ = 115°

To find the x- and y-components of the velocity of part 2, we can use the given angle and speed:

vx2 = v2 * cos(θ)
vy2 = v2 * sin(θ)

Now, let's calculate the x- and y-components of the total momentum before the explosion:

Total momentum before explosion in the x-direction:
= (0.500 kg) * (10.0 m/s) + (0.750 kg) * (v2 * cos(115°))

Total momentum before explosion in the y-direction:
= (0.500 kg) * (12.0 m/s) + (0.750 kg) * (v2 * sin(115°))

Since the package is stationary before the explosion, the total momentum in both directions is zero:

Total momentum before explosion in the x-direction: 0 = (0.500 kg) * (10.0 m/s) + (0.750 kg) * (v2 * cos(115°))

Total momentum before explosion in the y-direction: 0 = (0.500 kg) * (12.0 m/s) + (0.750 kg) * (v2 * sin(115°))

Now, we can solve these equations to find the value of v2:

0 = (0.500 kg) * (10.0 m/s) + (0.750 kg) * (v2 * cos(115°))

0 = (0.500 kg) * (12.0 m/s) + (0.750 kg) * (v2 * sin(115°))

Solving these equations will give us the values of v2.

(a) The speed of part 3 can be found using the equation:

v3 = sqrt(v3x^2 + v3y^2)

(b) The direction of part 3 can be found using the equation:

θ3 = atan(v3y / v3x)

Substituting the values of v3x and v3y into these equations will give us the final answers to (a) and (b).

Note: To perform the calculations, you may need a scientific calculator or computational software.