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makkiz [27]
3 years ago
6

A panpipe is made of five pipes. The longest pipe is 25 centimeters long

Physics
1 answer:
Marysya12 [62]3 years ago
7 0
The pipe that produces the highest-frequency sound is the 5 cm long pipe. 

According to the Fundamental Principle, the length of the tube of the pipe and its frequency is inversely proportional. 

This means that the longer the tube, the lower the frequency and vice versa. Therefore, between the 25cm long pipe and the 5cm long pipe, the shorter pipe produces the highest-frequency sound.
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In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If t
Vitek1552 [10]

Answer:

So the ratio will be \frac{T_L}{T_H}=-0.171

Explanation:

We have given heat engine absorbs 450 joule from high temperature reservoir

So Q=450j

As the heat engine expels 290 j

So work done W = 290 J

We know that efficiency \eta =\frac{W}{Q}=\frac{290}{450}=0.6444

It is given that efficiency of the engine only 55 % of Carnot engine

So efficiency of Carnot engine =\frac{0.6444}{0.55}=1.171

Efficiency of Carnot engine is \eta =1-\frac{T_L}{T_H}

1.171 =1-\frac{T_L}{T_H}

\frac{T_L}{T_H}=-0.171

3 0
3 years ago
Read 2 more answers
An 4.0-kg fish pulled upward by a fisherman rises 1.9 m in 2.4 s, starting
mina [271]

Answer:

2.64N  

Explanation:

Force = mass * acceleration

Given

mass = 4kg

distance = 1.9m

Time t = 2.4s

Get the acceleration using the equation of motion

S = ut + 1/2at²

1.9 = 0 + 1/2a(2.4)²

1.9 = 5.76a/2

1.9 = 2.88a

a = 1.9/2.88

a = 0.66m/s²

Get the magnitude of the force

Force = 4 * 0.66

Force = 2.64N

Hence the net force acting on the fish is 2.64N  

5 0
3 years ago
A negative point charge q1 = 25 nC is located on the y axis at y = 0 and a positive point charge q2 = 10 nC is located at y =14
sergey [27]

Answer:

 y = 0.1 m

Explanation:

The electrical power for point loads is

         V = k \sum \frac{q_i}{r_i}k Sum qi / ri

in this case

         V = k (- \frac{q_1}{r_1 } + \frac{q_2}{r_2})

indicate that V = 0

        \frac{q_1}{r_1} = \frac{q_2}{r_2}

        r₂ = \frac{q_2}{q_1} r_1

the distance r1 is

         r₁ = y -0

the distance r2

         r₂ = 0.14 -y

we substitute

       

        0.14 - y = \frac{10}{25}  y

          y ( \frac{10}{25} + 1) = 0.14

          y 1.4 = 0.14

          y = 0.14 / 1.4

          y = 0.1 m

7 0
3 years ago
A ball travels with velocity given by Error converting from MathML to accessible text., with wind blowing in the direction given
KengaRu [80]

Answer: The magnitude of the velocity = 2/5 m/s

Explanation:

 In this question, the magnitude of the velocity is the product of the magnitude of the displacement vector and the magnitude of the component of the velocity that acts in the direction of displacement.

This will be a scalar projection of V onto X

Please find the attached files for the solution

5 0
3 years ago
The sound from a loud speaker has an intensity level of 80 db at a distance of 2.0m. Consider the speaker to be a point source,
Tamiku [17]

Answer:

2.83m

Explanation:

The information that we have is

Intensity at 2.0 m: I=80dB and r_{1}=2m

we need an intensity level of: I_{2}=40dB

thus, we are looking for the distance r_{2}.

which we can find with the law for intensity and distance:

(\frac{r_{2}}{r_{1}} )^2=\frac{I_{1}}{I_{2}}

we solve for r_{2}:

\frac{r_{2}}{r_{1}}=\sqrt{\frac{I_{1}}{I_{2}} }\\\\r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}} }

and we substitute the known values:

r_{2}=(2m)\sqrt{\frac{80dB}{40dB} }\\\\r_{2}=(2m)\sqrt{2}\\ r_{2}=2.83m

at a distance of 2.83m the intensity level is 40dB

5 0
3 years ago
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