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a person sitting in a parked car hears an approaching ambulance siren at a frequency f1. as it passes him and moves away, he hears a frequency f2. the actual frequency f of the source is (which one of the following)a. f > f1b. f < f2c.f= f2 - f1d. f = f2 + f1e. f2 < f < f1

Answers

Answer 1

Answer:

Answer:

e. f2 < f < f1

Explanation:

According to Doppler's Effect:

$(f_o)/(f_s) =(S+v_o)/(S-v_s)$ ......................................(1)

where:

$f_o\ \&\ f_s$ are observed frequency and source frequency respectively.

S = velocity of sound in the air from a stationary source

$v_o\ \&\ v_s$ are the velocity of the observer and the velocity of sound source with respect to a stationary frame of reference.

When the ambulance approaches a stationary observer

Here $v_o=0\$

Then eq. (1) becomes:

$(f)/(f1) =(S)/(S-v_s)$

Now, the value:

$(f1)/(f) =(S)/(S-v_s)>1$

$\therefore f<f1$

Now according to the given condition the source is moving away from the observer i.e. the velocity of the source is opposite to the velocity of sound with respect to the stationary observer.

Now the eq. (1) becomes

$(f2)/(f) =(S)/(S-(-v_s))$

∵the direction of motion of the source is away from the observer so a negative sign has been introduced.

Now, the value:

$(f2)/(f) =(S)/(S+v_s)<1$

$\therefore f>f2$

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Find the average speed of the electrons in a 1.0 cm diameter, copper power line, when it carries a current of 20 A.

Neptunium. In the fall of 2002, scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. Neptunium-237 has a density of 19.5 g/cm3. What would be the radius of a sphere of this material that has a critical mass?

Answers

To solve this problem it is necessary to apply the concepts related to density, such as the relationship between density and Volume.

The volume of a sphere can be expressed as

$V = (4)/(3) \pi r^3$

Here r is the radius of the sphere and V is the volume of Sphere

Using the expression of the density we know that

$\rho = (m)/(V) \rightarrow V = (m)/(\rho)$

The density is given as

$\rho = (19.5g/cm^3)((10^3kg/m^3)/(1g/cm^3))$

$\rho = 19.5*10^3kg/m^3$

Now replacing the mass given and the actual density we have that the volume is

$V = (60kg)/(19.5*10^3kg/m^3 )$

$V = 3.0769*10^(-3) m ^3$

The radius then is,

$V = (4)/(3) \pi r^3$

$r = \sqrt[3]{(3V)/(4\pi)}$

Replacing,

$r = \sqrt[3]{(3(3.0769*10^(-3)))/(4\pi)}$

The radius of a sphere made of this material that has a critical mass is 9.02 cm.

The data listed below are for mechanical waves. Which wave has the greatest energy?

Answers

theres no data listed below

Answer:

amplitude = 14 cm; wavelength = 7 cm; period = 12 seconds

Explanation:

An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Determine the speed at this time and the maximum height at which it reaches.

Answers

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2* 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+(at^2)/(2)$

$s=0+(2* 6^2)/(2)$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=(144)/(2* 9.8)=7.34 m$

$s+s_0=36+7.34=43.34 m$

An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe — an insulating sphere with a diameter of 4 m and charge density 3nC/m2 on its outside surface. The sphere "captures" the electron, which falls into a circular orbit. Required:
Find the radius and period of the orbit.

Answers

Answer:

r = 2,026 10⁹ m and T = 2.027 10⁴ s

Explanation:

For this exercise let's use Newton's second law

F = m a

where the force is electric

F = $k (q_1q_2)/(r^2)$

Acceleration is centripetal

a = v² / r

we substitute

$k (q_1q_2)/(r^2) = m (v^2)/(r)$

r = $k (q_1q_2)/(m \ v^2)$ (1)

let's look for the charge in the insulating sphere

ρ = q₂ / V

q₂ = ρ V

the volume of the sphere is

v = 4/3 π r³

we substitute

q₂ = ρ $(4)/(3)$ π r³

q₂ = 3 10⁻⁹ $(4)/(3)$ π 4³

q₂ = 8.04 10⁻⁷ C

let's calculate the radius with equation 1

r = 9 10⁹ 1.6 10⁻¹⁹ 8.04 10⁻⁷ /(9.1 10⁻³¹ 628 10³)

r = 2,026 10⁹ m

this is the radius of the electron orbit around the charged sphere.

Since the orbit is circulating, the speed (speed modulus) is constant, we can use the uniform motion ratio

v = x / t

the distance traveled in a circle is

x = 2π r

In this case, time is the period

v = 2π r /T

T = 2π r /v

let's calculate

T = 2π 2,026 10⁹/628 103

T = 2.027 10⁴ s

Air contains 78.08% nitrogen, 20.095% oxygen, and 0.93% argon. calculate the partial pressure of oxygen if the total pressure of the air sample is 1.7 atm.a.

Answers

partial pressure in a mixture of two or more gases will be given by formula

$P_(partial)$ = mole fraction of gas * total pressure

now here mole fraction is same as percentage of gas in the mixture

Now mole fraction of oxygen is 0.20095 (20.095%)

now here pressure of oxygen in the mixture is given as

$P_(o_2) = 0.20095 * P_(total)$

$P_(o_2) = 0.20095 * 1.7$

$P_(o_2) = 0.342 atm$

so pressure due to oxygen in the mixture will be 0.342 atm

Answer:

20.095

Explanation:

An internal explosion breaks an object, initially at rest,intotwo pieces, one of which has 1.5 times the mass of the other.If
7500 J were released in the explosion, how much kinetic energydid
each piece acquire?

Answers

Answer:

4500 J and 3000 J

Explanation:

According to conservation of momentum

$0 = m_1 V_1 + m_2 V_2$

Given that m_2 = 1.5 m_1 , so

$V_1 = -1.5 V_2$

the kinetic energy of each piece is

$K_2= (1)/(2) m_2v_2^2$

$K_1= (1)/(2) m_1v_1^2$

substituting the value of V1 in the above equation

$K_1 = (1/2)( m_2 / 1.5 )( -1.5 V_2)^2 = 1.5 (1/2)m_2 V_2^2 = 1.5 K_2$

Given that

K_1 + k_2 = 7500 J

1.5 K_2 + K_2 = 7500

K_2 = 7500 / 2.5

= 3000 J

this is the KE of heavier mass

K_1 = 7500 - 3000 = 4500 J

this is the KE of lighter mass

Final answer:

The question is about finding the kinetic energy acquired by each of two pieces of an object following an internal explosion, using principles of conservation of energy and momentum in physics.

Explanation:

The student has asked about an internal explosion that breaks an object into two pieces with different masses, releasing a certain amount of kinetic energy in the process. This question involves applying the principle of conservation of energy and momentum to find the kinetic energy acquired by each piece post-explosion.

Assuming piece 1 has a mass of m and piece 2 has a mass of 1.5m, the total mass of the system is 2.5m. Since 7500 J of energy was released in the explosion, to find the kinetic energy of each piece, we can use the fact that the total kinetic energy is equal to the energy released during the explosion. Let the kinetic energy of the smaller piece be K1 and of the larger piece be K2. Because the object was initially at rest and momentum must be conserved, the momenta of the two pieces must be equal and opposite. This relationship allows us to derive the ratio of the kinetic energies. We can solve for K1 and K2 proportionally. Finally, because the kinetic energy is a scalar quantity, adding the kinetic energies of the two pieces will equal the total energy released.

Learn more about Kinetic Energy Acquisition here:

brainly.com/question/28545352

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