could anyone determinate the period knowing that it performs 4000 vibrations in 0.5 minutes, Sorry my english is bad

A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certa

Aliun [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

b

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

c

The ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

Explanation:

Mathematically Frequency can be represented as

$F = \frac{v}{\lambda}$

Where $\lambda$ is the wavelength and v is the velocity

Now looking at the diagram we see that

For the first frequency we have

Let the wavelength be $\lambda_1 = \lambda$ , and the frequency $F_1 = F$

For the second frequency

Let the wavelength be $\lambda_2 = 2 \lambda$ , and the frequency $F_2 = \frac{F}{2}$

For the third frequency

Let the wavelength be $\lambda_3 = 2\lambda$ , and the frequency $F_3 = \frac{F}{2}$

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

So,

For the first frequency we have

$v_1 = \lambda_1 F_1 = \lambda F$

For the second frequency

$v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F$

For the third frequency

$v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F$

Hence

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

Mathematically angular speed can be represented as

$w = 2 \pi f$

For the first frequency we have

$w_1 = 2\pi F_1 = 2 \pi F$

For the second frequency

$w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2} = \pi F$

For the third frequency

$w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2} = \pi F$

Hence

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

$v = wr$

=> $r = \frac{v}{w}$

Since the linear velocity is constant we have that

$r \ \alpha \ \frac{1}{w}$

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

5 0

3 years ago

Assume an 8-kg bowling ball moving at 2m/s bounces off a spring at the same speed that it had before bouncing.What is the moment

aliya0001 [1]

Answer:-16kgm/s

Explanation:

-mv=-2×8=-16kgm/s

5 0

3 years ago

Only one of three balls A, B, and C carries a net charge q. The balls are made from conducting material and are identical. One o

Zarrin [17]

Answer:

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Explanation:

The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.

Let's apply this principle to our case.

One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.

qA = ½ q

qB = ½ q

qC = 0

The total charge is q

we make a second contact

If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half

qA = 1/2 (q / 2) = ¼ q

qC = ¼ q

qB = ½ q

At this point all spheres have a charge,

qA = ¼ q

qb = ½ q

qC = ¼ q

The total charge is q

Now let's contact spheres B and one of the other two

Q = ½ q + ¼ q = ¾ q

When splitting the charge

qB = ½ ¾ q = 3/8 q

qC = ½ ¾ q = 3/8 q

qA = ¼ q

The total charge is q

Note that the total load is always equal to q

Now let's analyze the given configuration

Let's look for the total load

Q = qA + QB + QC

Q = ½ q + 3/8 q + ¼ q

Q = 9/8 q

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

8 0

3 years ago

PLS HELP FIND THE AVREGE KENETIC ENERGY!!

ANEK [815]

Answer: if you gogle it it will tell you the awenser

lanation:

4 0

3 years ago

two skateboarders of mass 50 kg and 60 kg push each other with force 70N.what is the acceleration of each skaters.step by step.

Keith_Richards [23]

Answer:

f=ma

f=70N

m1=50kg

m2= 60kg

a1= f/m

a1= 70/50

a1= 1.4N/kg

a2= 70/60

a2=1.17N/kg

6 0

3 years ago