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1 year ago

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Mrs. Gonzalez is about to give birth and Mr. Gonzalez is rushing her to the hospital at a speed of 30.0 m/s. Witnessing the spee

ding car, Officer O'Malley jumps in his police car and turns on the siren, whose frequency is 800. Hz. If the officer chases after the Gonzalez' car with a speed of 35.0 m/s, what frequency do the Gonzalezes hear as the officer approaches?

1 answer:

valina [46]1 year ago

3 0

Answer: The frequency = 1714.3Hz

Explanation: The solution can be achieved by using doppler effect formula.

Since the source is moving toward the observer, the velocity of the observer will be positive.

Please find the attached file for the solution

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A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°

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Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume $V_{1} = 2.00\ L$

Pressure $P_{1}= 1.00\ atm$

Pressure $P_{2}= 70.0\ kPa$

Temperature $T_{1}= 20.0°C = 293\ K$

Temperature $T_{2}= 7.00°C = 280\ K$

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Using gas law

$\dfrac{PV}{T}=\ Constant$

For both temperature,

$\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$

Put the value into the formula

$\dfrac{101.325\times2}{293}=\dfrac{70\times V_{2}}{280}$

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1 year ago

A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho

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Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

$80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)$

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$80=250-5x$

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$x=34 cm$

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1 year ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

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1 year ago