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ANTONII [103]
1 year ago
8

Q|C The speed of a one-dimensional compressional wave traveling along a thin copper rod is 3.56 km/s . The rod is given a sharp

hammer blow at one end. A listener at the far end of the rod hears the sound twice, transmitted through the metal and through air, with a time interval Δt between the two pulses.(d) Imagine that the copper rod is replaced by another material through which the speed of sound is v_r . What is the length of the rod in terms of t and v_r ?
Physics
1 answer:
Gekata [30.6K]1 year ago
8 0

The length of rod in terms v(r) and t is L = [  Δt / (1/343) - (1/v(r)) ].

3.56 km/s is the speed of a one-dimensional compressional wave moving along a thin copper rod.

At one end of the rod, a hard hammer strike is delivered. With a time interval of Δt between the two pulses, a listener at the other end of the rod hears the sound twice as it travels through the metal and the air.

The time interval is given by t = L/v.

The delay between pulses arrivals is:

Δt = L [(1/v(air)) - (1/v(copper))]

Now,

When the copper rod is swapped out for a different substance and the sound speed is measured as v(r).

The speed of air, v(air) = 343 m/s

Then,

L = [  Δt / (1/v(air)) - (1/v(r)) ]

L = [  Δt / (1/343) - (1/v(r)) ]

Here L is the length of the rod, Δt is in seconds and v(r) is the speed of sound in the rod.

Learn more about speed here:

brainly.com/question/13943409

#SPJ4

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Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

7 0
3 years ago
A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region
zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with  due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

K_{1} = K_{2} + W_{f}

Where:

K_{1}, K_{2} are the initial and final translational kinetic energies of the tobbogan, measured in joules.

W_{f} - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})

Where:

f - Friction force, measured in newtons.

\Delta s - Distance travelled by the toboggan in the rough region, measured in meters.

m - Mass of the toboggan, measured in kilograms.

v_{1}, v_{2} - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}

If m = 375\,kg, v_{1} = 4.50\,\frac{m}{s}, v_{2} = 1.20\,\frac{m}{s} and \Delta s = 5.40 \,m, then:

f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}

f = 653.125\,N

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%

\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%

\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%

\%K_{loss} = 92.889\,\%

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%

\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%

\%v_{loss} = 73.333\,\%

The speed of the toboggan is reduced in 73.333 %.

5 0
3 years ago
Imagine a simple pendulum swinging in an elevator. If the cable holding the elevator up was to snap, allowing the elevator to go
Lady_Fox [76]

One of the components that affect the period is gravity (the other is length). This gravity is basically the value of the effective acceleration that acts on the body due to gravity. When the elevator is over free fall, the effective gravity becomes zero. Mathematically this can be visualized as,

T = 2\pi \sqrt{\frac{l}{\dot{g_{eff}}}}

Since this value is zero, the period would tend to be infinite,

T \rightarrow \infty

Therefore the frequency that is inversely proportional to the period would be defined as

f = \frac{1}{T}

f = \frac{1}{\infty}

f \approx 0

In this way there is no frequency on the body which will not generate any oscillation on the body

5 0
3 years ago
Please help, my homework is due in a couple of days and I'm stuck. A plane is travelling 10.0 meters per second (this means the
ss7ja [257]

Answer:

  • < 25 m/s
  • triangle inequality
  • between north and east
  • 45° < angle < 60°

Explanation:

(a) Just as one-dimensional numbers add on a number line by putting them end-to-end, so two-dimensional numbers add on a coordinate plane the same way.

Here, we choose to let the positive y-axis represent North, and the positive x-axis, East. This is the way a map is conventionally oriented. The velocity of the plane is represented by a vector pointing north (up). Its length represents the magnitude of the velocity. Likewise, the wind is represented by a vector of length 15 pointing east (right). The sum of these is the hypotenuse of the triangle they form.

The magnitude of the sum can be found here using the Pythagorean theorem, but for the purpose of this question, you're not asked to find that.

Instead, you're asked to estimate whether it is more or less than 25 (m/s).

Your knowledge of the triangle inequality will tell you that the hypotenuse (resultant) must be shorter than the sum of the lengths of the sides of the triangle, hence must be less than 10+15 = 25.

__

(b) The triangle inequality says the resultant is less than the sum of the other two sides of the triangle.

__

(c) Since the wind is blowing the plane toward the east, but the plane is traveling toward the north, the resulting direction is somewhere between north and east.

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(d) "Somewhere between north and east" can be expressed as the inequality ...

  0° < angle < 90°

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Answer:

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Explanation:

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