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2 years ago

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The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end

s, how many resonances are between the two given ones, and if the pipe is closed at one end, how many resonances are between the two given ones?

1 answer:

Xelga [282]2 years ago

8 0

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency of pipe is given by

$f=\frac{nv}{2L}$

where L=length of Pipe

v=velocity of sound

$f=150\ Hz$ for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

$f=\frac{(2n+1)}{4L}\cdot v$

for n=0

$f_1=\frac{v}{4L}$

$f_1=150\ Hz$

for n=2

$f_2=\frac{5v}{4L}$

Thus there is one additional resonance corresponding to n=1 , between $f_1$ and $f_2$

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A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas

vodka [1.7K]

Answer:

$v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

$F=qE$

$F_p=F_e$

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

$m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE$

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

$\frac{m_ev_e^2}{2d}=qE$

we can equals these expressions for both proton and electron, because the forces qE are the same:

$\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

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2 years ago

The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha

saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm = 4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

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2 years ago

A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant

OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

$a_t = ar$

where;

a is the angular acceleration and

r is the radius of the circular path

$a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2$

Determine time of the rotation;

$\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\$

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

$a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2$

The magnitude of total linear acceleration is given by;

$a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2$

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

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2 years ago

Ice fishermen sit on top of frozen lakes in the winter and catch fish in the liquid water below through holes cut in the ice she

melomori [17]

Answer:

This is because below 4°c, water unlike other materials becomes less dense when it's temperature is further lowered.

Explanation:

Due to the unusual nature of water; at about 4°c, the behavior of the density of water in relation to its temperature reverses. This means that water becomes less dense as it becomes colder below 4°c. The colder parts therefore floats to the top of the water body while the warmer part sinks allowing the top to freeze and the remaining body below to remain in its liquid state.

The freezing of the top of the lake alone protects the remaining depth of water from freezing by acting as an insulator and preventing further heat loss from the water to the ambient space. If this had not been the case, and water froze all through, marine lives will freeze to death and it will be more difficult to melt the ice come the next summer.

This behavior is due to the hydrogen bonding of the water molecules.

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2 years ago

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then

Elan Coil [88]

Answer:

A) $W_{ff} =-744.12J$

B) $F_f=-W_{ff}*sin\theta /hy = 112.75N$

C) $F_{f2}=207.58N$

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s. </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement. </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder. </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

$W_{ff}=\Delta E$

$W_{ff}=1/2*m*vf^2-m*g*hy$

$W_{ff}=-744.12J$

For part B), we use our previous value for the work:

$W_{ff}=-F_f*(hy/sin\theta)$ Solving for friction force:

$F_f=-W_{ff}*sin\theta /hy$

$F_f=112.75N$

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

$Vf^2=Vo^2+2*a*d$

Solving for a:

$a=-3.844m/s^2$

Now, by dynamics:

$|F_f|=|m*a|$

$|F_f|=207.58N$

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2 years ago