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Dmitry_Shevchenko [17]
3 years ago
14

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x, t) = (5

.60 cm) sin [(0.0340 rad/cm) x] sin [(50.0 rad/s) t], where the origin is at the left end of the string, the x-axis is along the string and the y-axis is perpendicular to the string.
a) Draw a sketch that shows the standing wave pattern.
(b) Find the amplitude of the two travelling waved that make up this standing wave.
(c) What is the length of the string?
(d) Find the wavelength, frequency, period and speed of the travelling waves.
(e) Find the maximum transverse speed of a point on the string.
(f) What would be the equation y(x, t) for this string if it were vibrating in its eighth harmonic?
Physics
1 answer:
barxatty [35]3 years ago
4 0

Answer:

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t]y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t], where the origin is at the left end of the string, the x-axis is along the string, and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation y(x, t) for this string if it were vibrating in its eighth harmonic?

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tresset_1 [31]

So the initial velocity is 15 m/s, the final velocity is 0 since it's at a complete stop and time is 10 seconds.  Therefore:

Acceleration=\frac{v_{final}-v_{initial}}{t_{final}-t_{initial}} =\frac{0m/s-15m/s}{10s-0s} =-1.5\frac{m}{s^2}

Therefore, the acceleration is -1.5 m/s^2.  The reason it's negative is due to the fact that the vector is going against it's original movement since it's decelerating.

7 0
3 years ago
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Find the work w1 done on the block by the force of magnitude f1 = 95.0 n as the block moves from xi = -5.00 cm to xf = 4.00 cm .
Vlad [161]
<h3><u>Answer;</u></h3>

= 8.55 Joules

<h3><u>Explanation;</u></h3>

Work done is the product of force and the distance moved by an object.

Work done = Force × distance

Force = 95 Newtons

Distance = X2 -X1

               = 4 - (-5)

               = 9 cm

Thus;

work done = 95 × 9/100

                  <u>= 8.55 Joules </u>

5 0
3 years ago
A block of mass 0.08 kg is pushed against a spring with spring constant k=31 N/m. The spring is compressed 0.15 meters from its
ELEN [110]

Answer:

1.11 meters

Explanation:

As the spring is compressed, elastic potential energy is built up in the spring. The total elastic potential energy can be found using the following formula

Ep = 1/2 x k x s²          

where k = 31 N/m  (spring constant)

s = 0.15 m  (compression)

Ep = 3.4875 J

When the block of mass is released, the elastic potential energy (Ep) is converted to kinetic energy (Ek). From this we can find the initial velocity of the mass of block after release

Ek = 1/2 x m x u²    

   

where Ek = Ep = 3.4875J

m = 0.08 kg  (mass of block)

u = unknown (initial velocity)

u = 2.9526 m/s

Now that we know the initial velocity we need to find the deceleration of the mass of block due to friction. We will first find the force of friction from the following formula

F = ∪ x m x g          

where F = unknown (frictional force)

∪ = 0.4   (coefficient of friction)

m = 0.08 kg   (mass of block)

g = 9.81 m/s² (acceleration due to gravity)

F = 0.31392 N

From this force we calculate the deceleration based on the following formula

F = m x a                  

where F = 0.31392   (frictional force)

m = 0.08 kg   (mass of block)

a = unknown  (acceleration)

a = -3.924 m/s²      -

*the negative sign is due to this value being deceleration

Now to find the total distance traveled we use the equation for motion

v² = u² + 2as            

where  v = 0 (final velocity)

u = 2.9526 m/s (initial velocity

a = -3.924 m/s² (deceleration due to friction)

s = unknown (distance traveled)

s = 1.11 meters

3 0
3 years ago
Describe liquids and GASSES IN TERMS OF THERE VOLUME AND SHAPES
antiseptic1488 [7]

Answer:

Explanation:

liquids have definite volume

liquids do not have definite shape. The take the shape of the container in which they are kept.

gases do not have definite volume.

gases do not have definite shape. They take the shape of the container in which they are kept.

Hope this helps

plz mark as barinliest!!!!!!

Stay safe!

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klio [65]

Explanation:

Let us assume that the separation of plate be equal to d and the area of plates is 9 \times 10^{-4} m^{2}. As the capacitance of capacitor is given as follows.

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It is known that the dielectric strength of air is as follows.

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Expression for maximum potential difference is that the capacitor can with stand is as follows.

                       dV = E × d

And, maximum charge that can be placed on the capacitor is as follows.

               Q = CV

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                   = 1.062 \times 10^{-8} C

or,                = 10.62 nC

Thus, we can conclude that charge on capacitor is 10.62 nC.

5 0
3 years ago
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