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Gennadij [26K]
3 years ago
9

Imagine you are waiting for a train to pass at a railroad crossing. Will the train whistle have a higher pitch as the train appr

oaches you or after it has passed you by?
Physics
1 answer:
frosja888 [35]3 years ago
7 0
THE DOPPLER EFFECT. Anyways, it would have a higher whistle as it approaches you, when it gets to you it only gets quieter because it leaves after. Think of a motorcycle going by, its loud coming to you then as it passes it gets quieter.
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-Plz Help Meh-<br> Calculate the acceleration of a car if the force is 450N and the mass is 1300kg.
Allisa [31]

Answer:

Explanation:

F=Ma

450=1300×a

Ans=0.346m/s²

4 0
3 years ago
Two Objects of the same size will always have the same mass” Is this statement correct?
attashe74 [19]

No, they won't, mass coincides with density and objects have different densities a one pound lead ball would be smaller than a one pound copper one.

6 0
3 years ago
Im stuck can someone help me?
mixer [17]
In 2.34 hours, the bus travels 16.34 km.

Average speed = (distance covered) / (time to cover the distance).

Average speed = (16.34 km) / (2.34 hours) = 6.983 km/hr
3 0
3 years ago
Which of the following is NOT an example of kinetic energy being converted to potential energy?
Pani-rosa [81]

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6 0
3 years ago
A potential difference of 300 volts is applied to a 2.0-µf capacitor and an 8.0-µf capacitor connected in series. (a) What are t
zubka84 [21]

Answer:

a) Q1=Q2=480μC   V1=240V   V2=60V

b) Q1=96μC   Q2=384μC   V1=V2=48V

c) Q1=Q2=0C    V1=V2=0V

Explanation:

Let C1 = 2μC  and  C2=8μC

For part (a) of this problem, we know that charge in a series circuit, is the same in C1 and C2. Having this in mind, we can calculate equivalent capacitance first:

C=\frac{1}{\frac{1}{C1} +\frac{1}{C2} } = 1.6\mu C

Q_{T} = V_{T}*C_{T} = 480\mu C

V1 = \frac{Q1}{C1}=240V

V2 = \frac{Q2}{C2}=60V

For part (b), the capacitors are in parallel now. In this condition, the voltage is the same for both capacitors:

V1=V2   So,  \frac{Q1}{C1} = \frac{Q2}{C2}

Total charge is the same calculated for part (a), so:

\frac{Qt - Q2}{C1} = \frac{Q2}{C2}   Solving for Q2:

Q2 = 384μC    Q1 = 96μC.

Therefore:

V1=V2=48V

For part (c), both capacitors would discharge, since their total voltage of 300V would by applied to a wire (R=0Ω). There would flow a huge amount of current for a short period of time, and capacitors would be completely discharged. Q1=Q2=0C  V1=V2=0V

3 0
3 years ago
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