The tendency for an object at rest to remain at rest

Flauer [41]

<u>The frequency of the sound</u> determines the pitch .The higher the frequency , the more high pitched it will sound. if the frequency is very low, it will sound very low pitched.

This is affected by motion of a sound source because if the sound source moves faster, this makes the frequency higher( since frequency is the number of waves/revolutions/whatever per second)

So , if for example , the sound source makes 4 vibrations or whatever in a second , it will have a frequency of 4Hz , but if a sound source makes 8 vibrations or whatever in a second , it will have a frequency of 8Hz . And since pitch is determined by frequency , the sound pitch with a frequency of 8Hz will have a higher pitch than sound source with a frequency of 4Hz.

3 0

3 years ago

Read 2 more answers

A truck is traveling at a speed of 50 km/h. What happens when the truck slows down?

MaRussiya [10]

Answer:

it loses its momentum. In simpler ways it is really slow

Explanation:

3 0

2 years ago

A mass of 0.75 kilograms is attached to a spring/mass oscillator. A force of 5 newtons is required to stretch the spring 0.5 met

zlopas [31]

Answer:

b > 66.41 kg/s

Explanation:

The spring force F = -kx, where k = spring constant, the damping force f = -bv. The net force F' = F + f

F + f = ma

-kx - bv = ma

-kx -bdx/dt = md²x/dt².

Re-arranging the equation, we have

So, md²x/dt² + bdx/dt + kx = 0

Dividing through by m, we have

d²x/dt² + (b/m)dx/dt + (k/m)x = 0

This is a second-order differential equation. The characteristic equation is thus,

D² + (b/m)D + (k/m) = 0

Using the quadratic formula, we find D.

$D = \frac{-(b/m) +/- \sqrt{(b/m)^{2} - 4k/m} }{2}$

For an overdamped system,

$(b/m)^{2} - 4k/m} > 0$

$(b/m)^{2} > 4k/m}\\(b/m) > \sqrt{4k/m}} \\(b/m) > 2\sqrt{k/m}} \\b > 2\sqrt{km}}$

Now, k = F/x. Since the weight of the object causes the spring to stretch a distance of 0.5 m, k = mg/x where m = mass of object = 0.75 kg, g = 9.8 m/s² and x = x₀ =0.5 m.

Substituting k = mg/x into the inequality for b, we have

b > 2√{(mg/x₀)m}

b > 2√{(m²g/x₀)}

b > 2m√{g/x₀)}

b > 2 × 0.75 kg√{9.8 m/s²/0.5 m)}

b > 1.5 kg√{19.6/s²)}

b > 1.5 kg × 4.427/s

b > 66.41 kg/s

6 0

2 years ago

Identify the relationship between kinetic energy and gravitational potential energy for the cyclist at each position

sasho [114]

Answer:

follows are the responses to the given question:

Explanation:

The kinetic energy of every object moving. $PK=m \times v^2$ any entity lifted against the strength of gravity stores elastic potential of gravity.

$PE = height \times mass \times g$

Its total power of an independent device stays constant underneath the Mass conservation on Energy and it is, the kinetic energy plus potential energy is just like a fixed $(KE+PE=Standard)$ . So, if KE improves, it is valid that PE declines.

If the PE is now at least then KE has been at the highest. It is also valid that KE is reduced as PE is increased as well as the maximum PE, the minimum KE.

5 0

3 years ago

During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.