Which Model represents a lithium isotope? Use your periodic table for this one! Answer:-D

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.

Explanation:

Given:

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

A red cart has a mass of 4 kg and a velocity of 5 m/s. There is a 2-kg blue cart that is parked and not moving, thus its velocit

IRISSAK [1]

Answer:dam hold up

Explanation:

5 0

3 years ago

Neon, helium, methane, krypton, xenon, hydrogen, and ozone make up & what of the atmosphere

Alinara [238K]

Part of the upper atmosphere

6 0

3 years ago

A ball thrown straight up into the air is found to be moving at 6.79 m/s after falling 1.87 m below its release point. Find the

MaRussiya [10]

Answer:

3.07 m/s

Explanation:

We know that from kinematics equation

$v^{2}=u^{2}+2as$ and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity