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slamgirl [31]
3 years ago
10

You are standing on a street corner and you hear an ambulance siren. The pitch of the siren seems to be getting higher. What do

you expect will happen next?
A. The pitch of the siren will change unpredictably as it approaches you.
B. The siren will get louder and higher as the ambulance moves towards you.
C. The siren will eventually fade as the ambulance moves away from you.
D. The pitch of the siren will continue to get higher, but you will not see the ambulance because it is moving away.
Physics
2 answers:
Ira Lisetskai [31]3 years ago
8 0

The answer is B

I know this because if the siren seems to be getting higher that means that the Ambulance is coming towards you. Which is was answer b is saying.


Hope this helps please make me brainliest.

Gre4nikov [31]3 years ago
4 0

B.The siren will get louder and higher as the ambulance moves towards you


Hope this helped !

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HELP I WILL GIVE BRAINLYEST!!
zavuch27 [327]

Answer:

A-B-C

Explanation:

Depends if you have good taste.

8 0
3 years ago
3.5 g of a hydrocarbon fuel is burned in a vessel that contains 250. grams of water initially at 25.00 C. After the combustion,
kherson [118]

Answer:

3.) 463 J/g

Explanation:

Heat given by the fuel is used to increase the temperature of the water

so it is given as

Q = ms\Delta T

here we will have

m = 250 g

\Delta T = 26.55 - 25

now we also know that

s = 4186

so we have

Q = (0.250)(4186)(26.55 - 25)

Q = 1622.075 J

so 3.5 g fuel gives above energy

so per gram of fuel will have total energy given as

Q = \frac{1622.075}{3.5}

Q = 463.45 J/g

6 0
3 years ago
14) What is the velocity of a wave 5.35 x 10 cm long with a frequency of 16kHz
JulijaS [17]

Answer:

wavelength= 5.35×10cm=0.535m

f= 16000hz

v= wavelength × f= 0.535×16000=8560m/sec

6 0
3 years ago
Suppose a car manufacturer tested its cars for front-en4 collisions by hauling them up on a crane and dropping then; from a cert
Brrunno [24]

Answer:

a

Generally from third equation of motion we have that

v^2 =  u^2 + 2a[s_i - s_f]

Here v is the final speed of the car

u is the initial speed of the car which is zero

s_i is the initial position of the car which is certain height H

s_i is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

v^2 = 0 + 2g[H - 0]

=> v  =  \sqrt{ 2 g H}

b

H  =  9.86 \  m

Explanation:

Generally from third equation of motion we have that

v^2 =  u^2 + 2a[s_i - s_f]

Here v is the final speed of the car

u is the initial speed of the car which is zero

s_i is the initial position of the car which is certain height H

s_i is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

v^2 = 0 + 2g[H - 0]

=> v  =  \sqrt{ 2 g H}

When v  = 50 \  km/h = \frac{50 *1000}{3600} = 13.9 \  m/s we have that

13.9  =  \sqrt{ 2 g H}

=> H  =  \frac{13.9^2}{2 *  9.8}

=> H  =  9.86 \  m

6 0
3 years ago
The propeller of a WW2 fighter plane is 2.35 m in diameter
Studentka2010 [4]
A) 1 rev = 2π rad. Using this ratio, you can find the rad/s: 1160 rev/min x 2π rad/rev x 1 min/60 s = 121.5 rad/s

b) You can find linear speed from angular speed using this equation (note the radius is half the diameter given in the question): v = ωr = 121.5 rad/s x 1.175 m = 142.8 m/s

c) You can find centripetal acceleration using this equation: a = v^2/r = (142.8 m/s)^2 / 1.175 m = 17 355 m/s^2
8 0
3 years ago
Read 2 more answers
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