Which statement correctly describes the location, charge, and mass of the

From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequ

Alexandra [31]

Approximately 15 m/s is the speed of the car.

<u>Explanation:</u>

<u>Given:</u>

speed of sound - 343 m/s

You detect a frequency that is 0.959 times as small as the frequency emitted by the car when it is stationary. So, it can be written as,

$f^{\prime}=0.959 f$

$\frac{f^{\prime}}{f}=0.959$

If there is relative movement between an observer and source, the frequency heard by an observer differs from the actual frequency of the source. This changed frequency is called the apparent frequency. This variation in frequency of sound wave due to motion is called the Doppler shift (Doppler effect). In general,

$f^{\prime}=\left(\frac{v+v_{0}}{v-v_{s}}\right) \times f$

Where,

$f^'$ - Observed frequency

f – Actual frequency

v – Velocity of sound waves

$v_0$ – Velocity of observer

$v_s$ - velocity of source

When source moves away from an observer at rest ( $v_{0} = 0$ ), the equation would be

$f^{\prime}=\left(\frac{v}{v-\left(-v_{s}\right)}\right) \times f$

$f^{\prime}=\left(\frac{v}{v+v_{s}}\right) \times f$

$\frac{f^{\prime}}{f}=\left(\frac{v}{v+v_{s}}\right)$

By substituting the known values, we get

$0.959=\left(\frac{343}{343+v_{s}}\right)$

$0.959\left(343+v_{s}\right)=343$

$0.959(343)+0.959\left(v_{s}\right)=343$

$328.937+0.959 v_{s}=343$

$0.959 v_{s}=343-328.937=14.063$

$v_{s}=\frac{14.063}{0.959}=14.66 \mathrm{m} / \mathrm{s}$

Approximately 15 m/s is the speed of the car.

3 0

3 years ago

The hydrogen atom, changing from its first excited state to its lowest energy state, emits light with a wavelength of 122 nm. Th

tatyana61 [14]

Answer:

true b and c

Explanation:

n the electromechanical transitions of the atoms the relationship must be fulfilled

$\frac{x}{\lambda }$ = R (1 / nf - 1 / no²)

where for the final state nf = 1 giving in the case of hydrogen the Lymma series whose smallest wavelength is lam = 122 nm with nf = 1 and there are a series of spectral lines for each value of n of the final state

in the case of sodium so well it has a transition from an excited state to the kiss state (bad)

Now let's review the different proposals

a) False. The electronic potential for sodium is much lower than for hydrognosia

b) True

c) True

d) true

8 0

3 years ago

The pressure on a balloon holding 356 mL of an ideal gas is increased from 267 torr to 1.00 atm. What is the new volume of the b

Natalija [7]

Answer:

124.52 mL

Explanation:

from Boyle's Law,

PV = P'V' ................... Equation 1

Where P = Initial pressure of the gas, V = Initial volume of the gas, P' = Final pressure of the gas, V' = Final volume of the gas.

make V' the subject of the equation.

V' = PV/P'............. Equation 2

Given: P = 267 torr = (267×0.00131) = 0.34977 atm, V = 356 mL, P' = 1 atm

Substitute into equation 2

V' = (0.34977×356)/1

V' = 124.52 mL.

Hence the new volume of the balloon = 124.52 mL

4 0

3 years ago

Round the following off to required number of significant figures.

Stells [14]

Answer:

140

hope this helps

have a good day :)

Explanation:

4 0

2 years ago

Read 2 more answers

Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative

SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

Explanation:

Given there are three blocks of masses $M_{a}$ , $M_{b}$ and $M_{c}$ (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *a ................(equation 1)

Also it is given that $M_{a}$ does not move with respect to $M_{c}$ , which gives tension T is exerted on pulley by $M_{a}$ only, Hence tension T is

T = $M_{a}$ *a ..........(equation 2)

There is also also tension exerted by $M_{b}$ . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = $M_{b}$ $\sqrt{a^{2} +g^{2} }$ ................(equation 3)