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cupoosta [38]
3 years ago
13

Which one of the following frequencies of a wave in the air can be heard as an audible sound by the human ear?

Physics
1 answer:
Setler [38]3 years ago
6 0

the answer is 1,000 A.

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A parked car's horn (belonging to a musician) emits a concert A of frequency 440 on a day when the speed of sound is 342 m/s. Yo
julsineya [31]

Answer:

19.08 m/s

Explanation:

f = actual frequency emitted by the parked car's horn = 440 Hz

V = speed of sound = 342 m/s

f' = frequency of the horn observed by you = 466 Hz

v = speed of your car moving towards the parked car = ?

frequency of the horn observed by you is given as

f' = \frac{Vf}{V - v}

466 = \frac{(342)(440)}{342 - v}

v = 19.08 m/s

3 0
3 years ago
Finish A 16 N force is applied to an object and 96 J of work is done. How far was the object moved?
Lina20 [59]

Answer:

\boxed {\boxed {\sf 6 \ meters}}

Explanation:

Work is the product of force and distance.

W=F*d

We know that 96 Joules of work were done and a 16 Newton force was applied to the object.

  • W= 96 J
  • F= 16 N

Substitute the values into the formula.

96 \  J= 16 \ N * d

First, let's convert the units. This will make cancelling units easier later in the problem. 1 Joule (J) is equal to 1 Newton meter (N*m), so the work of 96 Joules equals 96 Newton meters.

96 \ N*m= 16 \ N * d

Now, solve for distance by isolating the variable, d. It is being multiplied by 16 Newtons and the inverse of multiplication is division. Divide both sides of the equation by 16 N.

\frac {96 \ N*m}{16 \ N}= \frac{16 \ N *d}{16 \ N}

\frac {96 \ N*m}{16 \ N}=d

The units of Newtons cancel.

\frac {96}{16} \ m = d

6 \ m = d

The object moved a distance of <u>6 meters.</u>

3 0
3 years ago
Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect
Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

Where Vo = initial voltage

e = natural logarithm = 2.718

For the first capacitor 2.50 µF,

V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF 

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

4 0
3 years ago
A material that has high resistance to the flow of electric current is called an electric ______
lyudmila [28]
A material that has high resistance to the flow of electric current is called an electric resistor
3 0
3 years ago
Read 2 more answers
Question 4 How much time does it take to walk 8 km north at a velocity of 3.8 km/h?​
IrinaVladis [17]

Given parameters:

Displacement = 8km

Velocity  = 3.8km/h

Unknown:

time  = ?

Solution:

Velocity is displacement divided by time.

  Velocity  = \frac{displacement}{time}  

      Displacement  = velocity x time

Input the parameters:

              8  = 3.8  x time

 Time  = \frac{8}{3.8}   = 2.1s

The time taken is 2.1s

6 0
3 years ago
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