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Morgarella [4.7K]

3 years ago

7

g Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the bu

zzer over the side while the balloon is rising at constant speed, what can you conclude about the sound you hear as the buzzer falls toward the ground

1 answer:

dlinn [17]3 years ago

8 0

Answer:

the observed frequency will reduce but the wavelength will increase

Explanation:

As we know

fo = fs (v/(v-vs))

fo = observed frequency

vs = velocity of source

As per this equation,

When an observer moves away from the stationary source, the observed frequency reduces. Since the observer in the balloon is moving away from the source which itself is moving in opposite direction, the observed frequency will reduce.

Since wavelength = V/fs . The source frequency is unchanged but the velocity is increasing as it is moving in downward direction. Hence, the wavelength will increase

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A 920-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

GrogVix [38]

Answer:21.45 m/s

Explanation:

Given

Mass of sport car=920 kg

Mass of SUV=2300 kg

distance to which both car skid is 2.4 m

coefficient of friction ( $\mu$ )=0.8

Let u be the initial velocity of both car at the starting of skidding

and they finally come to zero velocity

$v^2-u^2=2as$

$acceleration=\mu g=0.8\times 9.8=7.84 m/s^2$

s=2.4 m

$0-(u)^2=2\times (-7.84)\times 2.4$

u=6.13 m/s

so before colliding sport car must be travelling at a speed of

$920\times v=(920+2300)\times 6.13$ (conserving momentum)

v=21.45 m/s

7 0

3 years ago

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

6 0

3 years ago

A car initially traveling at 27.2 m/s undergoes a constant negative acceleration of magnitude 1.90 m/s2 after its brakes are app

zubka84 [21]

Answer:

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

Explanation:

We can use the following equation:

$\omega_{f}^{2}=\omega_{i}^{2}-2\alpha \Delta \theta$ (1)

The angular acceleration is:

$a_{tan}=\alpha R$

$\alpha=\frac{1.9}{0.325}$

$\alpha=5.85\: rad/s^{2}$

and the initial angular velocity is:

$\omega_{i}=\frac{v}{R}$

$\omega_{i}=\frac{27.2}{0.325}$

$\omega_{i}=83.69\: rad/s$

Now, using equation (1) we can find the revolutions of the tire.

$0=83.69^{2}-2*25.85 \Delta \theta$

$\Delta \theta=135.47\: rad$

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

I hope it helps you!

6 0

3 years ago

The tropic of cancer is to the tropic of capricorn as the arctic circle is to the

Y_Kistochka [10]

Answer:

Antarctic Circle

Explanation:

The Tropic of Cancer, which is also referred to as the Northern Tropic, is the most northerly circle of latitude on Earth at which the Sun can be directly overhead. This occurs on the June solstice, when the Northern Hemisphere is tilted toward the Sun to its maximum extent.

Tropic of Capricorn Is it Southern Hemisphere counterpart, marking the most southerly position at which the Sun can be directly overhead.

7 0

3 years ago

Any two application of gravity

Brrunno [24]

Answer:

Well the definition of an application is the act of putting to a special use or purpose so lam assuming that you want specific uses that scientists make of gravity in their work.

Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.

The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.

Hope that helps.

Explanation:

5 0

3 years ago