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2 years ago

Help with the two questions above? Correct answers?

Physics

1 answer:

LenKa [72]2 years ago

7 0

(6) first choice: the frequency appears higher and wavelength is shorter.

The car approaches a stationary observer and so the sound will appear to have shorter wavelength. This creates an effect of its siren to sound with higher frequency than it would do if both were stationary.

(7) The Doppler formula for frequency in the case of a stationary observer and source approaching it is as follows:

$f_O = \frac{v}{v-v_s}\cdot f= \frac{343\frac{m}{s}}{(343-25)\frac{m}{s}}\cdot 400Hz \approx 431Hz$

The wavelength is then

$\lambda = \frac{343\frac{m}{s}}{431Hz}\approx 0.80 m$

The third choice "0.80m; 431Hz" is correct

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Which are properties of both a gas and a plasma? Check all that apply.

wlad13 [49]

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3 years ago

Read 2 more answers

An object, with mass 32 kg and speed 26 m/s relative to an observer, explodes into two pieces, one 5 times as massive as the oth

Brut [27]

Answer:

$\Delta K = 2164.053\,J$

Explanation:

Let consider the observer as an inertial reference frame. The object is modelled after the Principle of Momentum Conservation:

$(32\,kg)\cdot (26\,\frac{m}{s} ) = (5.333\,kg)\cdot (0\,\frac{m}{s} )+(26.665\,kg )\cdot v$

The speed of the more massive piece is:

$v = 31.202\,\frac{m}{s}$

The kinetic energy added to the system is:

$\Delta K = \frac{1}{2}\cdot [(5.333\,kg)\cdot (0\,\frac{m}{s} )^{2}+(26.665\,kg )\cdot (31.202\,\frac{m}{s} )^{2}]-\frac{1}{2}\cdot (32\,kg)\cdot (26\,\frac{m}{s} )^{2}$

$\Delta K = 2164.053\,J$

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2 years ago

A 60-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the

Mariulka [41]

Answer:

part (a). 176580 J

part (b). 197381 J

Explanation:

Given,

Density of the chain = $\rho\ =\ 10\ kg/m.$
Length of the chain = L = 60 m
Acceleration due to gravity = g = 9.81 $m/s^2$

part (a)

Let dy be the small element of the chain at a distance of 'y' from the ground.

mass of the small element of the chain = $\rho dy$

Work done due to the small element,

$dw\ =\ \rho g (60\ -\ y)dy\\$

Total work done to wind the entire chain = w

$w\ =\ \displaystyle\int_{0}^{L} \rho g(60\ -\ y)dy\\\Rightarrow w\ =\ \rho g\left |(60y\ -\ \dfrac{y^2}{2})\ \right |_{0}^{60}\\\Rightarrow w\ =\ 10\times 9.81\times (60\times 60\ -\ \dfrac{60^2}{2})\\\Rightarrow w\ =\ 176580\ J$