An imaginary cubical surface of side L has its edges parallel to the x-, y- and z-axes, one corner at the point x = 0, y = 0, z

Response:

A=0.199

Clarification:

We know that  

Mass of spring=m=450 g=

Where 1 kg=1000 g

To determine the amplitude of oscillations.

Energy for oscillator=

Where =Angular frequency

A=Amplitude

Using the formula

Therefore, the amplitude of oscillation=A=0.199

We start by finding the angle of inclination with the sine function,

sin θ = 1 m / 4 m

θ = 14.48°

Next, we compute the work done by the movers using the following formula:

W = Fnet * d

We need to first determine Fnet. It is the weight force minus the frictional force.

Fnet = m g sinθ – μ m g cosθ

Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48

Fnet = 84.526 N

The work done is therefore:

W = 84.526 N * 4 m

<span>W = 338.10 J</span>

Answer:

130 m/s (to two significant figures)

Explanation:

In projectile motion, the launching velocity and launch angle help to determine both the horizontal and vertical velocity components.

u represents the initial projectile velocity = 150 m/s

uₓ = u cos θ = 150 cos 30° = 129.9 m/s

uᵧ = u sin θ = 150 sin 30° = 75.0 m/s

A projectile's motion can be viewed as made up of independent vertical and horizontal elements.

The vertical motion is affected by gravitational acceleration (which pulls down on the projectile), altering the vertical velocity component due to this acting force.

Conversely, there is no acting force in the horizontal direction, which means the horizontal component maintains a steady velocity throughout the projectile's flight.

Thus, at t = 4 s, the horizontal component of the projectile's speed remains equal to the initial horizontal velocity component.

At t = 4 s, the horizontal component of velocity is uₓ = u cos θ = 150 cos 30° = 129.9 m/s ≈ 130 m/s