A quantity y is to be determined from the equation y=(px)/q^2

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the ot

s344n2d4d5 [400]

Answer:

The length of the longer pipe is L = 2.30 m

Explanation:

Given that:

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.

How long is the other pipe?

From above;

The formula for the frequency of open ended pipes can be expressed as:

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

where n = 1 ( since half wavelength exist between those two pipes)

v = 343 m/s and L = 2.08 m

Thus, the shorter pipe produces a frequency of :

$f = \dfrac{1*343}{2*2.08}$

$f = \dfrac{343}{4.16}$

$f =82.45 \ Hz$

Also; we know that the beat frequency was given as 8.0 Hz

Then,

The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz

The lower frequency of the longer pipe = 74.45 Hz

Finally;

From the above equation; make Length L the subject of the formula. Then,

The length of the longer pipe is L = $\dfrac{nv}{2f}$

The length of the longer pipe is L = $\dfrac{1*343}{2*74.45}$

The length of the longer pipe is L = $\dfrac{343}{148.9}$

The length of the longer pipe is L = 2.30 m

6 0

2 years ago

Determine the length of a copper wire that has a resistance of 0.172 ? and cross-sectional area of 7.85 × 10-5 m2. The resistivi

KonstantinChe [14]

Answer:

Length of copper wire, l = 785 meters

Explanation:

Given that,

Resistance of the copper wire, R = 0.172 ohms

Area of cross section, $A=7.85\times 10^{-5}\ m^2$

Resistivity of copper, $\rho=1.72\times 10^{-8}\ \Omega-m$

The resistance of a wire is given by :

$R=\rho\dfrac{l}{A}$

$l=\dfrac{RA}{\rho}$

$l=\dfrac{0.172\ \Omega\times 7.85\times 10^{-5}\ m^2}{1.72\times 10^{-8}\ \Omega-m}$

l = 785 meters

So, the length of the copper wire is 785 meters. Hence, this is the required solution.

8 0

2 years ago

The number that is used to show the value of one currency compared to another is called the __________. A. trade rate B. currenc

Brrunno [24]

Answer:

c is the answer

Explanation:

8 0

2 years ago

Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C , which you find totally unacceptable. You

AleksAgata [21]

Answer:

$\eta=0.5074\ or\ 50.74\%$

Explanation:

Considering the density & specific heat capacity of coffee to be equal to that of water.

GIVEN:

density $\rho=1\ g.mL^{-1}$

specific heat $c=4.186\ J.g^{-1}.K^{-1}$

mass of coffee, $m=200\times 1=200\ g$

initial temperature of coffee, $T_i=30^{\circ}C$

final temperature of coffee, $T_f=60^{\circ}C$

power rating of oven, $P=1100\ W$

time taken to reach the final temperature, $t=35\ s$

Heat released by the coffee to come to 60°C:

$Q=m.c.\Delta T$

$Q=200\times 4.186\times 30$

$Q=[tex]\eta=\frac{25116}{49500}$ \ J[/tex]

Now the energy used by the oven in the given time:

$E=P.t$

$E=1100\times 45$

$E=49500\ J$

Now the efficiency:

$\eta=\frac{Q}{E}$

$\eta=0.5074\ or\ 50.74\%$

8 0

2 years ago

Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms o

ipn [44]

Answer:

U = 0.5 * k *(x + d - h_max)^2 + m*g*h_max

Explanation:

Given:

- The extension in spring @ equilibrium = x m

- The spring constant = k

- The amount of distance pulled down = d

- mass of the toy = m

Find:

- The total mechanical energy E_top at the top position h_max in terms of the available variables.

Solution:

- First we need to determine the types of Energy that are in play:

- The Elastic potential Energy E_p in a spring is given:

E_p: 0.5 * k * (ext)

- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:

ext = Equilibrium + distance pulled - h_max = (x + d - h_max)

Hence, the elastic potential energy will be:

E_p = 0.5 * k *(x + d - h_max)^2

- The gravitational potential energy E_g is given by:

E_g = m*g*h_max

Where, bottom most position is taken as reference (datum).

- The kinetic Energy E_k is given by:

E_k = 0.5*m*v_top^2

- Since we know that the maximum height is reached when velocity is zero

Hence, E_k = 0.5*m*0^2 = 0.

The total Energy of the system U is sum of all energies and play:

U = E_p + E_k + E_g

U = 0.5 * k *(x + d - h_max)^2 + m*g*h_max

8 0

2 years ago