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Aloiza [94]
3 years ago
15

Two cars, one in front of the other, are traveling down the highway at 53.7 m/s. The car behind sounds its horn, which has a fre

quency of 536 Hz. What is the frequency heard by the driver of the lead car
Physics
1 answer:
Vadim26 [7]3 years ago
4 0

Answer:

Explanation:

The source is approaching the observer which is moving away from the source , so applying Doppler's effect

n = n_o\frac{V-V_O}{V-V_S}

where n is apparent frequency , n₀ is original frequency , V₀ is velocity of observer , V_S is velocity of source , V is velocity of sound

Putting the given values

= n = 536\frac{340-53.7}{340-53.7 }

= 536

So apparent frequency will be same as original frequency .

ie 536 .

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A 22kg Accelerates at a rate of 2.3 m/s. What is the magnitude of the net force acting on the bike?
Tcecarenko [31]

magnitude of the net force = mass x acceleraton

                                             = 22 x 2.3

                                             =50.6 N

7 0
3 years ago
15. Find the speed of a disc of radius 0.5 meters and mass 2-kg at the base of the incline. The disc starts at rest and rolls do
mote1985 [20]

Explanation:

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7 0
3 years ago
A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th
Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

E=-\frac{d}{dt}(BACOS\alpha )\\

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using \pi r^{2} \\ where r is the radius of 5cm or 0.05m

A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\

since we no that the angle is at 0^{0}

we determine the magnitude of the magnetic filed

B=0.5e^{-t} \\t=2s

E=-(0.5e^{-2} * 0.00785)

E=-0.000532v\\

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A

I=2.6mA

6 0
3 years ago
A source produces 20 crests and 20 troughs in 4 seconds. The second crest is 3 cm away from the first crest.Calculate :
sineoko [7]

Answer:

Solution given:

No of waves[N] =20crests & 20 troughs

=20waves

Time[T]=4seconds

distance[d]=3cm=0.03m

Now

<u>Wave</u><u> </u><u>length</u><u>=</u>3cm=3 × {10}^{ - 2}m

<u>Frequency</u>=\frac{No of waves}{time}

=\frac{20}{4}=5Hertz

and

Wave speed:wave length×frequency=3 × {10}^{ - 2}m×5=1.5 × {10}^{ - 1} \tt{ {ms}^{ - 1}}.

3 0
2 years ago
a furnace supplies 28kW of thermal power at 300C to an engine and exhausts waste energy at 20C. At the very best, how much work
DochEvi [55]

Answer:

The amount of work we could expect to get out of the system per second = 28,000J/s

Explanation:

Given the power supplied to the system as  28kW;

Energy = power / time

At very best, the amount of work we could expect to get out of the system per second = 28,000 W / 1 second =  28,000J/s

Therefore, for a a furnace which supplies 28kW of thermal power at 300C to an engine and exhausts waste energy at 20C.

At the very best, the amount of work we could expect to get out of the system per second = 28,000J/s

6 0
3 years ago
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