A baseball is hit high into the upper bleachers of left field. Over its entire flight the work done by gravity and the work done

The correct answer is -10.

The fundamental equation is derived from Mr. Planck: E=h \nu, where h is Planck’s constant and ν is the frequency. This relationship describes the energy per photon at a specific frequency. Although a wavelength is provided, it can easily be converted to frequency using the equation: c= lambda / nu, where c denotes the speed of light; λ (lambda) is the wavelength; and ν is the frequency. Once the energy of a photon with a wavelength of 550nm is determined, it will show how many photons are needed to gather 10^-18J. Remember to pay attention to the units.

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 result to the query is 0.25 ohms

Explanation:

R = u x/A.......1

where u represents the resistivity of the

rod, A is the cross-sectional area, and x denotes

the length of the rod.

Let R* represent the resistance across the adjacent sections of the rod.

Then, R* = u1/4.......2

By comparing equation 1 with equation 2, we find that

R* = 1/4

which equals 0.25 ohms.