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Arisa [49]
3 years ago
14

A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear

a frequency of 65 Hz. What is the frequency of the sound emitted by the horn
Physics
1 answer:
Talja [164]3 years ago
4 0

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

f_o = f_s\frac {(v+vs)}{v} and f_o = f_s\frac {(v-vs)}{v} when moving towards observer

With f_o of 76 then taking speed in air as 343 m/s we have

76 = f_s\times\frac {(343-vs)}{343}

f_s=\frac {343\times 76}{343-v_s}

Similarly, with f_o of 65 we have

65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}

Now

f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}

v_s=27.76 m/s

Substituting the above into  any of the first two equations then we obtain

f_s=70.07 Hz

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A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for
faltersainse [42]

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

5 0
3 years ago
I’m gym class you run 22m horizontal then climb a rope vertically for 4.8m. What is the direction angle of your total displaceme
Sliva [168]

The answer is: 12.30 degrees.

To solve this problem you must apply the proccedure shown below:

1. You run 22 meters horizontally and then you climb a rope vertically for 4.8 meters.

2. So. let's call the angle \alpha, therefore, you must find it using tan^{-1}, as following:

tan^{-1}(\alpha)=\frac{opposite}{adjacent} \\opposite=4.8\\adjacent=22

3. Substitute values and solve for the angle:

tan^{-1}(\alpha)=\frac{4.8}{22}\\\alpha=12.30degrees

5 0
3 years ago
A falling object with a mass of 2.55 kg encounters 4.0 N of air resistance. What is the acceleration of the object? (Choose up t
mel-nik [20]

The acceleration of the falling object with a mass of 2.55 kg that encounters 4.0 N of air resistance is 1.57m/s².

<h3>How to calculate acceleration?</h3>

Acceleration is change of velocity with respect to time (can include deceleration or changing direction).

The acceleration of a body can be calculated by using the following expression;

a = F/m

Where;

  • a = acceleration (m/s²)
  • F = force (N)
  • m = mass (kg)

According to this question, a mass of 2.55 kg encounters 4.0 N of air resistance. The acceleration of the object can be calculated as follows:

a = 4N ÷ 2.55kg

a = 1.57m/s²

Therefore, 1.57m/s² is the acceleration of the object.

Learn more about acceleration at: brainly.com/question/28767690

#SPJ1

4 0
1 year ago
What is the velocity for an object at rest?
dem82 [27]

Answer:

e

Explanation:

An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force.

7 0
3 years ago
Read 2 more answers
A 1200-kg car accelerates it’s speed from 4 m/s to 10 m/s in 3 seconds. Find the car average speed , the car acceleration , the
dsp73

Answer:

Given: mass 1200kg

initial velocity: 4m/s

finial velocity: 10 m/s

time 3 sec

then

speed; initial velocity + final velocity/2

4+10/3

: 4.66m/s2

8 0
3 years ago
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