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Sliva [168]
2 years ago
8

An object is released from rest near and above Earth’s surface from a distance of 10m. After applying the appropriate kinematic

equation, a student predicts that it will take 1.43s for the object to reach the ground with a speed of 14.3m/s . After performing the experiment, it is found that the object reaches the ground after a time of 3.2s. How should the student determine the actual speed of the object when it reaches the ground? Assume that the acceleration of the object is constant as it falls.
Physics
1 answer:
Damm [24]2 years ago
4 0

Answer:

v_y = 12.54 m/s

Explanation:

Given:

- Initial vertical distance y_o = 10 m

- Initial velocity v_y,o  = 0 m/s

- The acceleration of object in air = a_y

- The actual time taken to reach ground t = 3.2 s

Find:

- Determine the actual speed of the object when it reaches the ground?

Solution:

- Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:

                             y = y_o + v_y,o*t + 0.5*a_y*t^2

                             0 = 10 + 0 + 0.5*a_y*(3.2)^2

                             a_y = - 20 / (3.2)^2 = 1.953125 m/s^2

- Use the principle of conservation of total energy of system:

                             E_p - W_f = E_k

Where,                  E_p = m*g*y_o

                             W_f = m*a_y*(y_i - y_f)      ..... Effects of air resistance

                             E_k = 0.5*m*v_y^2

Hence,                  m*g*y_o - m*a_y*(y_i - y_f) = 0.5*m*v_y^2

                             g*(10) - (1.953125)*(10) = 0.5*v_y^2

                             v_y = sqrt (157.1375)

                            v_y = 12.54 m/s

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Two different conducting objects, initially separated, are connected by a conducting wire. When the wire is removed, can you con
Alex Ar [27]

The two objects after (b) are at the same potential

Explanation:

When the two conducting objects are connected together, charge will be transferred from one object ot another until the two objects reach the same potential.

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8 0
2 years ago
Lucy has three sources of sound that produce pure tones with wavelengths of 60cm, 100cm, and 124cm.
denis23 [38]

Answer:

a) We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength

b) There is resonance for the lengths 25 and 75 cm

c) Resonance occurs for tubes with length 31 and 93 cm

Explanation:

To find the length of the tube that has resonance we must find the natural frequencies of the tubes, for this at the point that the tube is closed we have a node and the open point we have a belly; in this case the fundamental wave is

              λ = 4L

The next resonance called first harmonic    λ₃ = 4L / 3

The next fifth harmonic resonance               λ₅ = 4L / 5,

WE see that the general form is                    λ ₙ= 4L / n          n = 1, 3, 5 ...

Let's use these expressions for our problem

Let's start with the shortest wavelength.

a) Lam = 60 cm

Let's look for the tube length that this harmonica gives

               L = λ n / 4

To find the shortest tube length n = 1

               L = 60 1/4

              L = 15 cm

For n = 3

              L = 60 3/4

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For n = 5

              L = 60 5/4

              L = 75 cm

For n = 7

             L = 60 7/4

             L = 105cm

We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength, in different harmonics 1, 3 and 5

.b) λ = 100 cm

For n = 1

         L = 100 1/4

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For n = 3

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For n = 5

       L = 100 5/4

      L = 125 cm

There is resonance for the lengths 25 and 75 cm in the fundamental and third ammonium frequency

c) λ=  124 cm

       L = 124 1/4

       L = 31 cm

For the second resonance

      L = 124 3/4

      L = 93 cm

Resonance occurs for tubes with length 31 and 93 cm in the fundamental harmonics and third harmonics

8 0
2 years ago
Help me plz.<br>Show workings​
ddd [48]
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The answer is 0.8Ns
3 0
2 years ago
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