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expeople1 [14]
2 years ago
6

A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s

(a) If the cap is removed, what is the new fundamental frequency of the tube
Physics
1 answer:
sergey [27]2 years ago
8 0

Answer:

Y = V / f      where Y equals wavelength

4 Y1 = V / f1       for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2   for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1      dividing equations

f2 = 2 f1  

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

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Radiation with a wavelength of 238 nm shines on a metal surface and ejects electrons that have a maximum speedof 3.75 X 105 m/s.
Ulleksa [173]
The Answer is e) - Gold.
4 0
2 years ago
If it requires 3.0 J of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re
-BARSIC- [3]

Answer:

Work done = 13605.44

Explanation:

Data provided in the question:

For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J

The relation between Energy (U) and the elongation (s) is given as:

U = \frac{1}{2}kx^2   ................(1)

where,

k is the spring constant

on substituting the valeus in the above equation, we get

3.0 = \frac{1}{2}k\times0.021^2

or

k = 13605.44 N/m

now

for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m

using the equation 1, we have

U = \frac{1}{2}\times13605.44\times (0.062)^2

or

U = 26.149 J

Also,

Work done = change in energy

or

W = 26.149 - 3.0 = 23.149 J

4 0
2 years ago
What constant acceleration (in ft/s2) is required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds
SIZIF [17.4K]

Answer:

6.746 ft/s^2

Explanation:

v(t)=50

v(0)=27

t=5/3600 = 1/720 hours

v(t)-v(0)= a(t-0)

50-27= a(1/720)

a= 23*720= 16560 mi/h^2

16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2

4 0
3 years ago
A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.
Arturiano [62]

Answer:

a) \alpha=0.5117\ rad.s^{-2}

b) \theta=8.4\ rad

Explanation:

Given:

  • initial rotational speed of phonograph, \omega_i=0\ rad.s^{-1}
  • final rotational speed of phonograph, N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}
  • time taken for the acceleration, t=5.73\ s

a)

Now angular acceleration:

\alpha=\frac{\omega_f-\omega_i}{t}

\alpha=\frac{2.932-0}{5.73}

\alpha=0.5117\ rad.s^{-2}

b)

Using eq. of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2

\theta=8.4\ rad

5 0
2 years ago
A car travels east at a constant velocity. The net force on the car is:
Scrat [10]
F=ma
As velocity is constant, a=0
So, F=0

Hope this helps!
6 0
2 years ago
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