The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,

Response:

The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is ft/ sec

Clarification:

Provided:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec. 5 = 125 ft

x(5) = 10 ft/sec. 5 = 50 ft

At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

Now, let's calculate the derivative of distance in relation to time

By plugging in the values for h(t) and x(t) and simplifying, we arrive at,

= = ft / sec

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>

Inertia is universally present. It's important to note that inertia doesn't serve as the force keeping objects in circular paths; that role belongs to centripetal force, which is not always present. Centripetal force actively pulls objects towards the center of a circle. Both inertia and centripetal force contribute to the phenomenon of circular motion. Thank you, and enjoy your day;)

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

Final displacement equals +24484.5 nm. The path difference observed with red light (λ1 = 656.3 nm) with 158 bright spots can be represented as: Δr = 2d2 - 2d1 = 150λ1, leading to the equation 2d2 - 2d1 = 150λ1. Dividing both sides results in: d2 - d1 = 75λ1 - - - - eq1, with d1 being the distance from the beam splitter to the fixed mirror, and d2 indicating the position of the movable mirror when 158 bright spots appear. Then, with 114 bright spots, the path difference is Δr = 2d'2 - 2d1 = 114λ1, simplifying to 2d'2 - 2d1 = 114λ1. Subsequent division gives: d'2 - d1 = 57λ1, where d'2 is the revised position of the movable mirror. The displacement of the movable mirror, (d2 - d'2), can be calculated by subtracting eq2 from eq1, leading to: d2 - d'2 = 75λ1 - 57λ2, with λ1 equal to 656.3 nm and λ2 equal to 434.0 nm. Finally, this gives d2 - d'2 = 75(656.3) - 57(434), resulting in +24484.5 nm.