A girl and boy pull in opposite directions on a stuffed animal. The girl exerts a force of 3.5 N. The mass of the stuffed animal

Answer:

Consequently, we find that:

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

both wires share the same length

both wires carry an identical current

Given that wires are cylindrical in shape, we can determine the area for each case:

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

So we observe that:

a. vA = vB/4

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

Answer:

The distance before stopping is 1.52 m,

velocity is 4.0 m/s on the y-axis

Explanation:

The particle’s motion is two-dimensional due to acceleration along both the x and y axes; each axis can be addressed independently for calculations.

a) At the moment the particle starts to reverse, its velocity should be zero (Vfx = 0)

     Vfₓ = V₀ₓ + aₓ t  

     t = -  V₀ₓ/aₓ

     t = - 2.4/(-1.9)

     t=  1.26 s

When the particle stops, we calculate its position

     X1 = V₀ₓ t + ½ aₓ t²

     X1= 2.4 1.26 + ½ (-1.9) 1.26²

     X1= 1.52 m

At this point, the particle begins returning.

b) The velocity comprises both x and y components.

   For the x section, Vₓ = 0 m/s indicates a halt, but the y component retains a velocity

    Vfy= Voy + ay t

    Vfy= 0 + 3.2 1.26

    Vfy = 4.0 m/s

Thus, the velocity reads as

    V = (0 x^ + 4.0 y^) m/s

c) To graph the motion, we create a table listing position x and y at given time intervals; let's begin the calculations for equations

      X = V₀ₓ t+ ½  aₓ t²

      Y = Voy t + ½  ay t²

      X= 2.4 t + ½ (-1.9) t²

      Y= 0 + ½ 3.2 t²

      X= 2.4 t – 0.95 t²

      Y=   1.6 t²

With these equations, we construct two graphs for position against time, one for the x-axis and another for the y-axis

                       Chart for graphing

              Time (s)     x(m)            y(m)

                 0                0               0

                 0.5             0.960       0.4

          1       1.45          1.6

                 1.50      1.46      3.6

                2.00      1.00      6.4

Answer:

The provided documents adequately explain the situation. We'll identify the force exerted on the rope using Newton's third law, which underscores that this problem relates to the concept of equilibrium.

The outcome reveals that "The tension T1 in rope 1 matches the force F.

Explanation:

According to Newton's third law, for every action, there is an equal and opposite reaction.

Refer to the documents for a thorough solution and clarification of the issue.