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3 years ago

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Halley's comet orbits the sun roughly once every 76 years. It comes very close to the surface of the Sun on its closest approach

. Estimate the greatest distance of the comet from the sun. Is it still in the Solar System? What planet's orbit is nearest when it is out there?

1 answer:

Licemer1 [7]3 years ago

3 0

Answer:

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

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A seamount is an isolated land mass rising from the ocean floor.<br> a. True<br> b. False

iris [78.8K]

The answer is A, True.

7 0

3 years ago

At the race track, one race car starts its engine with a resulting intensity level of 104.0 dB at point P. Then 6 more cars star

dem82 [27]

To solve the problem it is necessary to apply the concepts related to sound intensity. The most common approach to sound intensity measurement is to use the decibel scale:

$\beta (dB) = 10log_{10}(\frac{I}{I_0})$

Where,

$I_0 = 1*10^{-12}$ is a reference intensity. It is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz.

I = Sound intensity

Our values are given by,

$\beta = 104dB$

$\#Autos = 7$

For each auto the intensity would be,

$104 = 10log\frac{I}{1*10^{-12}}$

$10.4= log_{10} (\frac{I}{10^{-12}})$

$10^{10.4}*10^{-12}=I$

$I = 0.02511W/m^2$

Therefore the sound intesity for the 7 autos is

$I= 7* 0.02511$

$I = 0.1748W/m^2$

The sound level for the 7 cars in dB is

$\beta (dB) = 10log_{10}(\frac{0.1748}{1*10^{-12}})$

$\beta (dB) = 112.42dB$

8 0

3 years ago

If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the

Viktor [21]

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

6 0

3 years ago

An object is moving in a straight line along the y axis. As a function of time, its position is given by the equation y=3.0+4.8x

romanna [79]

Answer:

Explanation:

<u>Instant Velocity and Acceleration </u>

Give the position of an object as a function of time y(x), the instant velocity can be obtained by

$v(x)=y'(x)$

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

$a(x)=v'(x)=y''(x)$

We are given the function for y

$y(x)=3.0+4.8x +6.4x^2$

Note we have changed the last term to be quadratic, so the question has more sense.

The velocity is

$v(x)=y'(x)=4.8+12.8x$

And the acceleration is $a(x)=v'(x)=12.8$

5 0

3 years ago

A 1250 kg car, driving 7.39 m/s, runs into the back of a stationary 5380 kg truck. After the collision, the truck moves forward

Leno4ka [110]

Explanation:

Given that,

Mass of the car, m₁ = 1250 kg

Initial speed of the car, u₁ = 7.39 m/s

Mass of the truck, m₂ = 5380 kg

It is stationary, u₂ = 0

Final speed of the truck, v₂ = 2.3 m/s

Let v₁ is the final velocity of the car. Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$1250\times 7.39+5380\times 0=1250\times v_1+5380\times 2.3$

$v_1=-2.5\ m/s$

So, the final velocity of the car is 2.5 m/s but in opposite direction. Hence, this is the required solution.

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4 years ago