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sdas [7]
4 years ago
12

Church organs have a set of pipes with different lengths. With those different pipes organs can produce sounds over a wide range

of frequencies. If the lowest frequency produced by an organ is 30.4 Hz, and the highest frequency is 1.48 kHz, then what is the shortest possible wavelength of sound the organ can produce? Assume that the speed of sound is 331 m/s.
Physics
1 answer:
Paraphin [41]4 years ago
8 0

Answer: The shortest possible wavelength of sound the organ can produce is 0.224 m

Explanation:

To calculate the wavelength of light, we use the equation:

\lambda=\frac{c}{\nu}

where,

\lambda = wavelength of the light

c = speed of light = 331m/s

As wavelength and frequency follows inverse relation, shortest wavelength is produced by highest frequency.

\nu = highest frequency of light = 1.48kHz=1.48\times 10^3Hz=1.48\times 10^3s^{-1}       1Hz=1s^{-1}

\lambda=\frac{331m/s}{1.48\times 10^3s^{-1}}=0.224m

Thus the shortest possible wavelength of sound the organ can produce is 0.224 m

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An object with a mass of 9 kg weighs 14.4 N on the Moon. What is the
erma4kov [3.2K]

Answer:

1.6 m/s²

Explanation:

Weight equals mass times acceleration due to gravity.

F = mg

14.4 N = (9 kg) g

g = 1.6 m/s²

7 0
4 years ago
Read 2 more answers
Two children of mass 18 kg and 29 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the c
DedPeter [7]

Answer:

3.085 [m].

Explanation:

1) The rule:

m₁*g*l₁=m₂*g*l₂, where m₁ and l₁ - the mass and distance for the small child, m₂ and l₂ - for the big child;

2) according to the condigion l₁+l₂=5, then

3) it is possible to make up the system:

\left \{ {{l_1+l_2=5} \atop {m_1*l_1=m_2*l_2}} \right. \ = > \ \left \{ {{l_1=5-l_2} \atop {18*(5-l_2)=29*l_2}} \right. \ = > \ \left \{ {{l_1=\frac{145}{47}} \atop {l_2=\frac{90}{47}}} \right.

4) finally, l₁=145/47≈3.085 [m].

7 0
2 years ago
The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the
nataly862011 [7]

Answer with Explanation:

Let  rest mass m_0 at point P  at  distance x from center of the planet, along a line connecting the centers of planet and the moon.

Mass of moon=m

Distance between the center of moon and center of planet=D

Mass of planet=M

We are given that net force on an object will be zero

a.We have to derive an expression for x in terms of m, M and D.

We know that gravitational force=\frac{GmM}{r^2}

Distance of P from moon=D-x

F_m=Force applied on rest mass due to m

F_m=Force on rest mass due to mas M

F_M=F_m because net force is equal to 0.

F_m=F_M

\frac{Gm_0m}{(D-x)^2}=\frac{Gm_0M}{x^2}

\frac{m}{(D-x)^2}=\frac{M}{x^2}

\frac{x^2}{(D-x)^2}=\frac{M}{m}

\frac{x}{D-x}=\sqrt{\frac{M}{m}}

Let R=\sqrt{\frac{M}{m}}

Then, \frac{x}{D-x}=R

x=DR-xR

x+xR=DR

x(1+R)=DR

x=\frac{DR}{1+R}

b.We have to find the ratio R of the mass of the mass of the planet to the mass of the moon when x=\frac{2}{3}D

Net force is zero

F_m=F_M

\frac{Gm_0m}{(D-\frac{2}{3}D)^2}=\frac{Gm_0M}{\frac{4}{9}D^2}

\frac{m}{\frac{D^2}{9}}=\frac{9M}{4D^2}

\frac{M}{m}=4

Hence, the ratio R of the mass of the planet to the mass of the moon=4:1

8 0
4 years ago
A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant
PIT_PIT [208]

Answer:

9.96\cdot 10^{-10}J

Explanation:

The capacitance of the parallel-plate capacitor is given by

C=\epsilon_0 k \frac{A}{d}

where

ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity

k = 3.00 is the dielectric constant

A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2 is the area of the plates

d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F

Now we can calculate the energy of the capacitor, given by:

U=\frac{1}{2}CV^2

where

C is the capacitance

V = 15.0 V is the potential difference

Substituting,

U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J

4 0
3 years ago
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Answer: c. increased sensitivity to ADH

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a. a decline in the number of functional nephrons: With aging the loss of nephron occurs that can be detected by the age related decrease in the glomerular filteration rate.

b. a reduction in the GFR (glomerular filtration rate): The GFR tend to decline in older age even though there is no disease. These people are required to check with the GFR in future.

d. problems with the micturition reflex: With aging people experience problem of bladder control. This leads to leakage or incontinence of urine or urinary retention that is inability to empty the bladder.

e. loss of sphincter muscle tone: With age the sphincter tone may diminish. This results in loss of control and storage capacity. The rectal muscles or sphincter muscles get loose which lead to passage of stool before reaching the washroom.

6 0
3 years ago
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