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White raven [17]
2 years ago
5

A 90.0-m-long brass rod is struck at one end. A person at the other end hears two sounds as a result of two longitudinal waves,

one traveling in the metal rod and the other traveling in the air. What is the time interval between the two sounds? Take the speed of sound in air to be 344 m/s. Use 8600 kg/m3 for the density of brass and 9.00×109 Pa for Young's modulus of brass.
Physics
1 answer:
Zanzabum2 years ago
6 0

Answer:

0.174413400763 s

Explanation:

L = Length of rod = 90 m

Speed of sound in air = 344 m/s

\rho = Density of brass = 8600 kg/m³

Y = Young's modulus = 9\times 10^9\ Pa

Velocity of sound in a material is given by

v_b=\sqrt{\dfrac{Y}{\rho}}\\\Rightarrow v_b=\sqrt{\dfrac{9\times 10^9}{8600}}\\\Rightarrow v_b=1022.99150921\ m/s

Time taken in brass

t=\dfrac{L}{v_b}\\\Rightarrow t=\dfrac{90}{1022.99150921}\\\Rightarrow t=0.0879772697913\ s

In air

t=\dfrac{L}{v}\\\Rightarrow t=\dfrac{90}{343}\\\Rightarrow t=0.262390670554\ s

The time interval is 0.262390670554-0.0879772697913=0.174413400763\ s

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Zolol [24]

Answer: 7.66 m/s

Explanation:

This situation is related to free fall (vertical motion). Hence, this can be considered a one-dimension problem and we can use the following equation:

V^{2}=V_{o}^{2}+2gd

Where:

V is the final velocity of the egg at the asked height

V_{o}=0 m/s is the initial velocity of the egg

g=9.8 m/s^{2} is the acceleration due gravity

d=4 m- 1m= 3m is the distance at which the egg is from the nest, when it is 1 m from the ground

Isolating V:

V=\sqrt{V_{o}^{2}+2gd}

Substituting the known values:

V=\sqrt{2(9.8 m/s^{2})(3 m)}

V=7.66 m/s This is the final velocity of the egg

3 0
2 years ago
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00
mote1985 [20]

Answer:

1. 6.99x 10^-6V/m

2. 18m

Explanation:

See attached file

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2 years ago
A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?
OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

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A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
Semenov [28]

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
2 years ago
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
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