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ASHA 777 [7]
4 years ago
5

Parasaurolophus was a dinosaur whose distinguishing feature was a hollow crest on the head. The 1.5-m-long hollow tube in the cr

est had connections to the nose and throat, leading some investigators to hypothesize that the tube was a resonant chamber for vocalization. If you model the tube as an open-closed system, what are the first three resonant frequencies?
Physics
1 answer:
Oksi-84 [34.3K]4 years ago
5 0

Answer:

f1 = 58.3Hz, f2 = 175Hz, f3 = 291.6Hz

Explanation:

lets assume speed of sound is 350 m/s.

frequencies of a standing wave modes of an open-close tube of length L

fm = m(v/4L)

where m is 1,3,5,7......

and fm = mf1

where f1 = fundamental frequency

so therefore: f1 = 350 x 4 / 1.5

f1 = 58.3Hz

f2 = 3 x 58.3

f2 = 175Hz

f3 = 5 x 58.3

f3 = 291.6Hz

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A person has a mass of 60 kg. What is the person’s weight in Newtons and in pounds?
liubo4ka [24]

Answer:

137.2 in pounds and in Newton's it's 588.399

3 0
3 years ago
A car is moving from rest and attained a velocity of 80 m/s. Calculate the acceleration of the car after 5 s?
soldi70 [24.7K]

Answer:

400 m/s

Explanation:

If it is moving at a velocity of 80 m/s from rest, then after 5 s it would be at 400 m/s.

80 x 5 = 400

7 0
3 years ago
A block with a mass of 33.0 kg is pushed with a horizontal force of 150 N. The block moves at a constant speed across a level, r
lord [1]

The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.

<h3>What is the work done?</h3>

The work done is given by the use of the formula;

W = F * x

Where;

F = force applied

x = distance covered

W = 150 N *  4.95 m = 742.5 J

Now;

The coefficient of kinetic friction is given by;

μ = F/mg

μ = 150/ 33 * 9.8

μ = 0.46

Learn more about work done:brainly.com/question/13662169

#SPJ1

4 0
2 years ago
A +17 nc charge is located at the origin. what is the strength of the electric field at the position (x,y)=(5.0cm,0cm)? express
gavmur [86]
The working equation to be used for this is written below:

E = kQ/d²
where
E is the electric field
k is a constant equal to 8.99 x 10⁹ N m²/C²
Q is the charge
d is the distance


E = (8.99 x 10⁹ N m²/C²)(17×10⁻⁹ C)/(0.05 m)²
E = 61,132 N/C
7 0
4 years ago
An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su
s2008m [1.1K]

Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0
3 years ago
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