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Ivenika [448]
3 years ago
11

The mass of a string is 5.9 × 10-3 kg, and it is stretched so that the tension in it is 200 n. a transverse wave traveling on th

is string has a frequency of 300 hz and a wavelength of 0.76 m. what is the length of the string?
Physics
1 answer:
bagirrra123 [75]3 years ago
4 0

The velocity of the wave on the string is given by

v=\sqrt{\frac{T}{\frac{m}{L}}}  \\  v=\sqrt{\frac{TL}{m}}

Solving the above equation,

v^2=\frac{TL}{m} \\  L=\frac{v^2m}{T}

The frequency of the wave f=300 and wave length is 0.76

The velocity is v=(300)(0.76)=228

Substituting numerical values,

L=\frac{228^2(0.0059)}{200}\\ T=1.534

The length of the string is 1.534 m

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1 because I looked it up for sure

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Use Newton's laws to explain why a falling object dropped from a 57m tower accelerates initially but then reaches constant veloc
snow_lady [41]

Answer:

At the point of dropping the object, by Newton's first law due to gravitational force F_g = m × g, accelerates

By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)

As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out

The body then moves at a constant uniform motion down according to Newton's first law

Explanation:

At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g

As the speed of the object increases while the object is falling with the gravitational acceleration the rate at which the object cuts through layers of air which (by Newton's first law of motion, are at rest ) has some buoyancy effect also increases therefore, the object is constantly increasingly changing the momentum of the air which by Newton's second law results, at an high enough velocity, and by Newton's third law, in a force equal to the applied gravitational force

Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.

5 0
3 years ago
Jasmine is diving off a 3-meter springboard. her height in meters above the water when she is x meters horizontally from the end
olganol [36]

Answer:

5.25 m

Explanation:

Given;

The height equation h;

h=-x^2+3x+3

Where;

h = the height above water

x = horizontal distance from the end of the board

The maximum height is at h' = 0, when change in h with respect to change in x is equal to zero.

differentiating the equation h.

dh/dx = h' = -2x + 3 = 0

Solving for x;

2x = 3

x = 3/2

Substituting into the function h;

h max = -x^2+3x+3

h max = -(3/2)^2 + 3(3/2) +3 = -9/4 +9/2 +3 = 9/4 + 3 =

h max = 21/4 = 5.25 m

8 0
3 years ago
A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori
olasank [31]

Explanation:

The given data is as follows.

        k = 130 N/m,       \Delta x = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = \frac{1}{2}kx^{2}

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}

   \frac{1}{2} \times 130 \times (0.17)^{2} = \frac{1}{2} \times 2.8 \times v^{2}

          v^{2} = \frac{3.757}{2.8}

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = \mu \times mg

                                    = 0.15 \times 2.8 \times 9.8

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           \frac{1}{2}mv^{2} = F_{f} \times d

           \frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d

             1.8515 = 4.116 \times d

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

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