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tiny-mole [99]
3 years ago
6

What is the heart rate recorded a few minutes after completing a workout?

Physics
2 answers:
DochEvi [55]3 years ago
8 0
Correct answer should be C
Naily [24]3 years ago
5 0
The answer is D, recovery heart rate
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Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips
Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

f = \frac{v}{\lambda}

Then the relation between two different frequencies with same wavelength would be

\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}

\frac{f'}{f} = \frac{v'}{v}

The beat frequency heard when the two strings are sounded simultaneously is

f_{beat} = f-f'

f_{beat} = f(1-\frac{f'}{f})

f_{beat} = f(1-\frac{v'}{v})

We have the velocity of the transverse waves in stretched string as

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{200N}{\mu}}

And,

v' = \sqrt{\frac{196N}{\mu}}

Therefore the relation between the two is,

\frac{v'}{v} = \sqrt{\frac{192}{200}}

\frac{v'}{v} = \sqrt{0.96}

Finally substituting this value at the frequency beat equation we have

f_{beat} = 590(1-\sqrt{0-96})

f_{beat} = 11.92Hz

Therefore the beats per second are 11.92Hz

4 0
2 years ago
Let w(x)=3x-7.If w(x)=14, find x
dlinn [17]

Answer:

7

Explanation:

We are given:

    w(x) = 3x   -   7

     w(x)  = 14

The problem here entails us to solve for x;

To solve for x; equate the two expressions:

         

          3x  - 7  = 14

           3x  = 14 + 7

           3x  = 21  

             x = 7

So the value of x  = 7

6 0
3 years ago
A net force of 10.0 N causes an object to accelerate at 2.00m/s^2. What is the mass of the object?
madam [21]

5.00kg

random = abcdefghijklmnopqrstuvwxyz


4 0
3 years ago
Determine the magnitude and direction of the resultant force of the following free body diagram.
Papessa [141]

Answer:

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

Explanation:

First, we must calculate the resultant force (\vec F), in newtons, by vectorial sum:

\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j} (1)

\vec F = 182.843\,\hat{i} + 356.048\,\hat{j}

Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:

\|\vec F\| = \sqrt{(482.843\,N)^{2}+(356.048\,N)^{2}}

\|\vec F\| \approx 599.923\,N

Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:

\theta = \tan^{-1}\left(\frac{356.048\,N}{482.843\,N} \right)

Where \theta is the direction of the resultant force, in sexagesimal degrees.

\theta \approx 36.405^{\circ}

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

4 0
2 years ago
An object traveling at a constant speed but with a changing direction is accelerating.
prohojiy [21]

Strange as it may seem, that's true. (choice 'a'.)

"Acceleration" doesn't mean "speeding up".  It means ANY change in
the speed or direction of motion.  So a car with the brakes applied
and slowing down, and a point on the rim of a bicycle wheel that's
turning at a constant rate, are both accelerating.

6 0
3 years ago
Read 2 more answers
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