Which option is an example of a physical property?

As an emergency vehicle approaches Bob and moves away from Jill, how does the actual frequency of the siren change? A) As an eme

Tamiku [17]

The correct answer is:
C) The actual frequency of the siren does not change despite appearances.

In fact, Bob will observe an increase in the apparent frequency as the emergency vehicle approaches him, while Jill will observe a decrease in the apparent frequency as the emergency vehicle moves away from him, because of the Doppler effect (the relative velocity between the observer and the source of the sound is changing), but this effect involves the apparent frequency, while the real frequency of the siren will remain the same.

7 0

3 years ago

Read 2 more answers

Bernice draws an oxygen atom. She draws a small circle for the nucleus. Inside of the circle, she draws plus signs for protons a

Lelu [443]

Answer

It should be A and C

Explanation:

because oxygen is number 8 in the periodic table of elements and has a atomic weight of 15.999 you use those numbers to figure out what is true between those.

The 8 for oxygen goes for the number of electrons and proton and to find neutrons u round the 15.999 up which now make it 16 and subtract it by the 8 now you have 8 protons, 8 neutrons, and 8 electrons

5 0

3 years ago

The graph shows the range of frequencies commonly heard by some animals.

Vsevolod [243]

Concept:

Frequency- It is defined as the number of oscillations occur in one second.

Its SI unit is Hertz (Hz)

Given: Produced sound vibrations is 18,500 cycles in 0.75 seconds

∵ In 0.75 second, produced sound has oscillations = 18,500 cycles

∴ In 1.0 second, produced sound has oscillations = (18,500 ÷ 0.75) Hz

The frequency of the sound will be ≈ 24,667 Hz

From the study of the given graph, only the animals (c) Cats, (b) Moths and (a) Bats can hear the produced sound because their upper audible frequency range is greater than 24,667 Hz.

3 0

3 years ago

Read 2 more answers

A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its

dlinn [17]

Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.

The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:

 70 km per second per megaparsec.

We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

 (Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =

 (150 million) x (1 / 3,262,000) x (70 km/sec) =

 3,219 km/sec in the direction away from us (rounded)

4 0

3 years ago

Brainliest please help tell me if am right if not correct me please

REY [17]

Answer:

See the answers below

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

First case

Vf = 6 [m/s]

Vo = 2 [m/s]

t = 2 [s]

$6=2+a*2\\4=2*a\\a=2[m/s^{2} ]$

Second case

Vf = 25 [m/s]

Vo = 5 [m/s]

a = 2 [m/s²]

$25=5+2*t\\t = 10 [s]$

Third case

Vo =4 [m/s]

a = 10 [m/s²]

t = 2 [s]

$v_{f}=4+10*2\\v_{f}=24 [m/s]$

Fourth Case

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

$v_{f}=5+8*10\\v_{f}=85 [m/s]$

Fifth case

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

$8=v_{o}+4*2\\v_{0}=8-8\\v_{o}=0$

8 0

3 years ago