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dangina [55]
3 years ago
10

A sound-producing object is moving away from an observer. The sound the observer hears will have a frequency that actually being

produced by the object.
the same as

higher than

lower than

need ASAP!!!
Physics
2 answers:
kozerog [31]3 years ago
6 0
Lower than, this is due to the doplar effect
padilas [110]3 years ago
5 0

Answer:

Lower than

Explanation:

According to Doppler’s effect, the observer will not hear the same frequency of sound if there is relative motion between the source and observer. When source is moving away from the observer the frequency perceived by the observer will be less than actual frequency as given by the following formula:

                                   f1 = (V/V+Vs)*f  

where,

f1 = frequency heard by the observer  

f = actual frequency  

V = velocity of sounds waves  

Vs = velocity of source

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Which describes the average velocity of an ant traveling at a constant speed
yanalaym [24]

Answer: A. The total displacement divided by the time and  C. The slope of the ant's displacement vs. time graph.

Explanation:

Hi! The question seems incomplete, but I found the options on the internt:

A. The total displacement divided by the time.

B. The slope of the ant's acceleration vs. time graph.

C. The slope of the ant's displacement vs. time graph.

D. The average acceleration divided by the time.

Now, since we know the ant is travelling at a constant speed, its average velocity V will be expressed by the following equation:

V=\frac{d}{t}

Where:

d is the ant's total displacement

t is the time it took to the ant to travel to the kitchen

Hence one of the correct options is: A. The total displacement divided by the time

On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is: C. The slope of the ant's displacement vs. time graph.

3 0
3 years ago
what is the resistance of a car light bulb that conducts 0.025A current when connected to a 12V car accumulator? is the current
Sophie [7]

One form of Ohm's Law says . . . . . Resistance = Voltage / Current .

R = V / I

R = (12 v) / (0.025 A)

R = (12 / 0.025) (V/I)

<em>R = 480 Ohms</em>

I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is.  (Floogle isn't sure either.)

If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.

If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.  

7 0
3 years ago
Using the same cost and time estimates, consider any trade-offs that SciTech may have to make to complete the project.
ch4aika [34]

Answer:  

Oracio is the most cost-effective choice because he would cost the least to complete the project. However, he would also take the longest amount of time. Camilla could complete the job the fastest, but she costs more than Oracio. SciTech will have to decide if it is more important to save money or complete the work quickly to meet the deadline.  

Hope this helps :)

4 0
3 years ago
Read 2 more answers
Carlos pushes a 3 kg box with a force of 9 newtons. The force of friction on the box is 3 newtons in the opposite direction. Wha
Ivenika [448]
The 'net' force acting on the box is (9 - 3) = 6 newtons
in the direction that Carlos is pushing.

Force = (mass) x (acceleration)

6 = (3) x (acceleration)

Divide each side by 3 :

<em>2 m/s² = acceleration</em>
4 0
3 years ago
Read 2 more answers
Two identical satellites are in orbit about the earth. One orbit has a radius r and the other 2r. The centripetal force on the s
velikii [3]

Answer:

the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

Explanation:

Mass of satellite, m

orbit radius of first, r1 = r

orbit radius of second, r2  = 2r

Centripetal force is given by

F= \frac{mv^{2}}{r}

Where v be the orbital velocity, which is given by

v=\sqrt{gr}

So, the centripetal force is given by

F= \frac{mgr}}{r}}=mg

where, g bet the acceleration due to gravity

g=\frac{GM}{r^{2}}

So, the centripetal force

F= \frac{GMm}}{r^{2}}}

Gravitational force on the satellite having larger orbit

F= \frac{GMm}{4r^{2}} .... (1)

Gravitational force on the satellite having smaller orbit

F'= \frac{GMm}{r^{2}} .... (2)

Comparing (1) and (2),

F' = 4 F

So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

8 0
3 years ago
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