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Vlad1618 [11]
2 years ago
10

A train is traveling at 30.0 m/sm/s relative to the ground in still air. The frequency of the note emitted by the train whistle

is 262 HzHz. The speed of sound in air should be taken as 344 m/sm/s .
Physics
1 answer:
vlada-n [284]2 years ago
6 0

frequency fapproach is heard by a passenger = 302.05 Hz

frequency frecede is heard by a passenger = 228.37433 Hz

Given :

speed of train to the ground = 30.0 m/s

frequency emitted by the train whistle = 262 Hz

speed of sound in air = 344 m/s

To Find :

(A) frequency fapproach (B) frequency frecede

Solution :

The frequency of approach is given by

f' = f <u>v + v(relative)</u>

v - v(air)

= 262x <u>344 + 18</u>

344 - 30

f' = 302.05 Hz

The frequency of approach is 302.05 Hz

The frequency of recede is given by

f' = f <u>v - v(relative)</u>

v + v(air)

= 262 x <u>344 - 18</u>

344 + 30

The frequency of recede is 228.37433 Hz

Learn more about Frequency here:

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When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature (37° C) by warmi
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Answer:

7.72 Liters

Explanation:

normal body temperature = T_body =37° C

temperature of ice water = T_ice =0°c

specfic heat of water = c_{water} =4186J/kg.°C

if the person drink 1 liter of cold water mass of water is = m = 1.0kg

heat lost by body is Qwater =mc_{water} ΔT

                                           = mc{water} ( T_ice - T_body)

                                             = 1.0×4186× (0 -37)

                                             = -154.882 ×10^3 J

here negative sign indicates the energy lost by body in metabolic process energy expended due to brisk - hour long walk is Q_{walk} = 286 kilocalories

            = 286×4186J

so number of liters of ice water have to drink is

n×Q_{water} =Q_{walk}                                                                        n= Q_{walk}/ Q_{water}

= 286×4186J/154.882×10^3 J

 = 7.72 Liters

7 0
3 years ago
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Balboa Park in San Diego has an outdoor organ. When the air temperature increases, the fundamental frequency of one of the organ
musickatia [10]

The fundamental frequency of one of the organ pipes will go up or increase.

When pressured air is forced into an organ pipe, it echoes at a particular pitch, generating the sound of the pipe organ. Each pipe has been adjusted to a particular pitch on the musical scale.

A musical instrument called an outdoor pipe organ is used to perform music. It produces some calming tones and has a really serene sound. The organ pipe produces the sound of the outdoor organ. The wavelength of the sound is also dependent on the length of the pipe. The fundamental frequency of one of the organ pipes will grow as the speed of the sound increases as the ambient air temperature rises.

The correct option is (c).

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6 0
1 year ago
A set of four capacitors are attached to a 12V battery in the circuit shown below. All capacitances are measured in milli-Farads
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The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is calculated as follows;

Capacitors in series;

1/Ct = 1/8 + 1/7.5

1/Ct = 0.25833

Ct = 3.87 mF

Capacitors is parallel;

Ct = 3.87 mF + 12 mF + 15 mF

Ct = 30.87 mF

Ct = 0.03087 F

<h3>Charge in each capacitor</h3>

Q = CV

Q = 0.03087 x 12

Q = 0.37 C

Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

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3 0
1 year ago
Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is
Zigmanuir [339]

Answer:

Part a)

I = 1.5 kg m^2

Part b)

I = 0.75 kg m^2

Part c)

I = 1.5 kg m^2

Explanation:

Part a)

Moment of inertia of the system about an axis passing through B and C is given as

I = mL^2 + mL^2 + m(0) + m(0)

I = 2mL^2

I = 2(3 kg)(0.50^2)

I = 1.5 kg m^2

Part b)

Moment of inertia of the system about an axis passing through A and C is given as

I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2

I = 2m\frac{L^2}{2}

I = (3 kg)(0.50^2)

I = 0.75 kg m^2

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2

I = 4m\frac{L^2}{2}

I = 2(3 kg)(0.50^2)

I = 1.5 kg m^2

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