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True [87]
3 years ago
8

A flock of geese is flying south for the winter. In order to maintain a “V” shape, each goose is flying at the same constant vel

ocity. Which goose will have the most momentum?
A) the goose with the least mass

B) the goose with the most mass

C) the goose that is behind all the others

D) the goose at the front of the "V" leading the way of the other geese
Physics
1 answer:
Novosadov [1.4K]3 years ago
3 0
B the goose with most mass
this is because momentum=mass x acceleration
so a larger mass will give a larger momentum ( acceleration stays constant)
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Answer:

object composition

Explanation:

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The loudness of sound refer to how loud or soft a sound seems to a listener
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Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each
AURORKA [14]

Answer: A (\sqrt{61},309.8°)

              B (2\sqrt{2}, 315°)

             C (3\sqrt{5}, 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

r=\sqrt{x^{2}+y^{2}} and \theta=tan^{-1}(\frac{y}{x})

For point A:

r=\sqrt{(-5)^{2}+6^{2}}

r=\sqrt{61}

\theta=tan^{-1}(\frac{6}{-5})

\theta=tan^{-1}(-1.2)

\theta=-50.2°

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

\theta=360-50.2

\theta= 309.8°

Polar coordinates for point A is (\sqrt{61}, 309.8°)

For point B:

r=\sqrt{2^{2}+(-2)^{2}}

r=\sqrt{8}

r=2\sqrt{2}

\theta=tan^{-1}(\frac{-2}{2} )

\theta=tan^{-1}(1)

\theta=-45°

Point B is in IV quadrant, so:

\theta=360-45

\theta= 315°

Polar coordinates for point B is (2\sqrt{2}, 315°)

For point C:

r=\sqrt{(-6)^{2}+(-3)^{2}}

r=\sqrt{45}

r=3\sqrt{5}

\theta=tan^{-1}(\frac{-3}{-6} )

\theta=tan^{-1}(0.5)

\theta= 26.56°

Polar coordinates for point C is (3\sqrt{5}, 26.56°)

3 0
3 years ago
You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 286.0 kPa when the ther
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Answer:

T_{2} = 606.69 K

Explanation:

In that the gas thermometer is a constant volume, it is satisfied that:

\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} }  

How the boiling water is under regular atmospheric pressure, then

T_{1} = 373 .15 K

Thus

\frac{286000}{373.15} = \frac{465000}{T_{2} }

T_{2} = 606.69 K

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If a particle's position is given by x=4-12t+3t^2, where t is in seconds and x is in meters, what is its velocity at t=1 second?
andreyandreev [35.5K]

Answer:

v = -6m/s

Explanation:

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