If 20 beats are produced within one second, which of the following frequencies could possibly be held by two sound waves traveling through a medium along a common path at the same time? A. 38 Hz and 63 Hz
B. 22 Hz and 42 Hz
C. 47 Hz and 55 Hz
D. 18 Hz and 26 Hz

Answers

Answer 1

Answer:

Answer:

B. 22 Hz and 42 Hz

Explanation:

The beat is an interference pattern which occurs when two waves of slightly different frequencies meet each other. The beat is heard as a periodic variation, whose frequency (called beat frequency) is equal to the absolute value of the difference between the frequencies of the individual waves:

$f_B = |f_1-f_2|$

In this problem, we are told that 20 beats per second are produced, so the beat frequency is

$f_B = 20 s^(-1) = 20 Hz$

Therefore, we must find the two waves which have a difference in frequency equal to this value, 20 Hz. Let's see each choice:

A. 38 Hz and 63 Hz --> $f_B = |38 Hz-63 Hz|=25 Hz$ --> NO

B. 22 Hz and 42 Hz --> $f_B = |22 Hz-42 Hz|=20 Hz$ --> YES

C. 47 Hz and 55 Hz --> $f_B = |47 Hz-55 Hz|=8 Hz$ --> NO

D. 18 Hz and 26 Hz --> $f_B = |18 Hz-26 Hz|=8 Hz$ --> NO

So, the correct choice is

B. 22 Hz and 42 Hz

Answer 2

Answer: The right answer for the question that is being asked and shown above is that: "B. 22 Hz and 42 Hz." If 20 beats are produced within one second, the frequencies that could possibly be held by two sound waves traveling through a medium along a common path at the same time is B. 22 Hz and 42 Hz

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A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 2.90 m/s. What is the average speed over the entire trip?

Answers

The Average Speed over the entire trip is 3.671 meters per second.

The person is moving in a straight line at a constant Speed. Let suppose that Distance between points A and B is $r$ , then we construct and use the Kinematic formulas for each stage of the travel to determine Times and Average Speed, based on the fact that Speed is inversely proportional to Time, we derive the resulting expression:

First Stage (from point A to point B):

$t_(1) = (r)/(v_(1))$ (1)

Second Stage (from point B to point A):

$t_(2) = (r)/(v_(2))$ (2)

Average Speed:

$t_(1)+t_(2) = (2\cdot r)/(\bar v)$ (3)

Where:

$t_(1)$ - Travelling time for the first stage, in seconds.

$t_(2)$ - Travelling time for the second stage, in seconds.

$r$ - Distance from A to B, in meters.

$v_(1)$ - Speed of the person in the first stage, in meters per second.

$v_(2)$ - Speed of the person in the second stage, in meters per second.

$\bar v$ - Average speed, in meters per second.

By applying (1) and (2) in (3), we derive an expression to determine the Average Speed:

$(r)/(v_(1)) + (r)/(v_(2)) = (2\cdot r)/(\bar v)$

$(v_(2)+v_(1))/(v_(1)\cdot v_(2)) = (2)/(\bar v)$

$\bar v = (2\cdot (v_(1)\cdot v_(2)))/(v_(2)+v_(1))$ (4)

If we know that $v_(1) = 5\,(m)/(s)$ and $v_(2) = 2.90\,(m)/(s)$ , the average speed over the entire trip is:

$\bar v = (2\cdot (v_(1)\cdot v_(2)))/(v_(2)+v_(1))$

$\bar v = (2\cdot \left(5\,(m)/(s) \right)\cdot \left(2.90\,(m)/(s) \right))/(2.90\,(m)/(s) + 5\,(m)/(s) )$

$\bar v = 3.671\,(m)/(s)$

The Average Speed over the entire trip is 3.671 meters per second.

Please see this question related to Average Speed for further details: brainly.com/question/19335778

This problem is a very interesting problem.
we know wacth:

Speed = distance / time.

Consider that:
distance from point A to point B=distance from point B to point A=d

We calculate the time to go from point A to point B.
time=distance / speed.
T₁=d / (5 m/s)

We calculate the time to go from point B to point A.
T₂=d / (2.9 m/s)

Therefore; the total time wil be: T₁+T₂
Total time=d/(5 m/s) + d / (2.9 m/s)
Least common multiple=(5 m/s)(2.9 m / s)=14.5 m²/s²
Total time=(2.9 m/s d + 5 m/s d)/ 14.5m²/s²
Total time=[(7.9 d) m/s] / (14.5 m²/s²)
Total time=7.9 d/ (14.5 m/s)

Now, the total distance will be =d+d=2d

Therefore:
Average speed=total distance / total time
Average speed=[(2d)] / [(7.9 d)/ (14.5 m/s)]
Average speed≈3.67 m/s.

Answer: the average speed will be ≈3.67 m/s.

What are three examples of circular motion?

Answers

Motion of fan, stone tied to a string and is being swung in circles and motion of electrons around nucleus.

What is the mechanical advantage of the lever shown below?

Answers

The mechanical advantage of the lever shown in the question is 1.27. Mechanical advantage is the force amplified by a mechanical tool. Thus, the correct option is B.

What is Mechanical advantage?

Mechanical advantage is the measure of the force amplification which is achieved by using a tool such as, a mechanical device or machine system. The mechanical device trades off input forces against the movement to obtain a desired amplification in the output force of the system.

Mechanical advantage of the lever is simply the ratio of the effort arm to the load arm. Effort arm is the distance from the pivot to the point of application of the force while the load arm is the distance of the lord from the pivot point.

Therefore, the effort arm is 0.27m while the load arm is 0.2 m. Mechanical advantage is calculated through the formula:

MA=effort arm/load arm

MA = 0.27/0.2

MA = 1.35 which is close to 1.27

Therefore, the correct option is B.

Learn more about Mechanical advantage here:

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Your question is incomplete, most probably the complete question is:

What is the mechanical advantage of the level shown below?

A. 0.79

B. 1.27

C. 2.6

D. 3.67

Answer:1.35

Explanation:

Suppose the fetus's ventricular wall moves back and forth in a pattern approximating simple harmonic motion with an amplitude of 1.7 mm and a frequency of 3.0 Hz. Find the maximum speed of the heart wall (in m/s) during this motion. Be careful of units!

Answers

Answer:

The maximum speed of the heart wall during this motion is 0.032 m/s.

Explanation:

Given that,

Amplitude of the simple harmonic motion, A = 1.7 mm = 0.0017 m

Frequency of the fetus's ventricular wall, f = 3 Hz

We need to find the maximum speed of the heart wall during this motion. The maximum speed of the object that is executing SHM is given by :

$v_(max)=A\omega$

$v_(max)=2\pi f A$

$v_(max)=2\pi * 3* 0.0017$

$v_(max)=0.032\ m/s$

So, the maximum speed of the heart wall during this motion is 0.032 m/s. Hence, this is the required solution.

PLEASE HELP ASAP!!!!!

Answers

Answer:

MOSE

Explanation:

Human activities can have an impact on natural disasters.

Answers

By definition, we have to:

The environmental impact are the consequences that some of the human activities produce on the environment.

The concept can be extended to the effects of a catastrophic natural phenomenon.

Human activities, such as fuel combustion and deforestation, have intensified the greenhouse effect, causing global warming.

It has been estimated that if emissions of greenhouse gases continue at the current rate, the temperature of the planet could exceed historical values in 2047, with harmful effects on ecosystems, biodiversity and endanger the livelihood of people on the planet.

Answer:

True.

Human activities can have an impact on natural disasters.

Human activities can have an impact on natural disasters is true

What is natural disasters?

The things humans do can affect natural disasters. Natural disasters are events that occur because of nature and its processes. However, human actions can also make these events happen more often, be worse, and affect more people in different ways.

Climate change is happening because of humans. We are releasing greenhouse gases when we burn fossil fuels and when we cut down trees. This is affecting the Earth's climate. This can cause more often and stronger natural disasters like hurricanes, droughts, heatwaves, and heavy rainfall.

Hence, Human activities can have an effect on natural disasters is a true statement.

Read more about natural disasters here:

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See full text below

Human activities can have an impact on natural disasters.true or false

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