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3 years ago

What is the number of complete wave cycles per unit time

Physics

1 answer:

kherson [118]3 years ago

3 0

If the unit time is the"second", then that's
the definition of the wave's <em>frequency</em>.

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3. A 0.35 kg puck slides across the ice with an average force of friction of 0.15 N acting on it. It slides 82 m before coming t

sattari [20]

Answer:

Work done is 12.3 J

Explanation:

We have,

Mass of puck, m = 0.35 kg

Force of friction acting on the puck when it slides is 0.15 N

Distance travelled by the puck is 82 m.

It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by

$W=Fd\\\\W=0.15\times 82\\\\W=12.3\ J$

The work done on the puck is 12.3 J.

5 0

4 years ago

Describe what happened. When was there more potential energy in the system?

nasty-shy [4]

Answer:

what is the full question

Explanation:

3 0

3 years ago

Analysis of the electrocardiogram can be revealed except ________

Lunna [17]

Answer:

when a person is not breathing

Explanation:

The electrocardiogram shows the cardiac action of the heart as a means of the sinusoidal waves. However, the waves have a different structure as they show the pumping phase, breathing and the resting phase of the heart. The waves continues to be displayed as long as there is systolic and diastolic pressure in the heart muscles. When there is no action, such as the cessation of brain activity, action ceases.

3 0

3 years ago

A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is

Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12 m

Let assume the tank have a thickness = dh

The radius of the tank by using the concept of similar triangle is :

$\dfrac{1}{r} = \dfrac{12}{h}$

$r = \dfrac{h}{12}$

The area of the tank = $\mathbf{\pi r^2}$

The area of the tank = $\mathbf{\pi( \dfrac{h}{12})^2}$

The area of the tank = $\mathbf{ \dfrac{\pi}{144}h^2}$

The volume of the tank is = area × thickness

= $\mathbf{ \dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\rho_ g * volume$

where;

$\rho_g$ = density of water ; which is given as 10000 N/m³

So;

Weight of the element = $\mathbf{ 10000 *\dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\mathbf{69.44 \ \pi \ h^2 \ dh}$

However; the work required to pump this water = weight × height rise

where the height rise = 12 - h

the work required to pump this water = $\mathbf{69.44 \ \pi \ h^2 \ dh}$ (12 - h)

the work required to pump this water = $\mathbf{69.44 \pi (12h^2-h^3)dh}$

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = $\mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi[ \frac{12h^3}{3}- \frac{h^4}{4}]^{12}}_0} }$

= $\mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}$

= $\mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}$

= $\mathbf{69.44 \pi*12^4 *\frac{1}{12}}$

= 376966.991 Joules

6 0

3 years ago

The wavelength of a sound wave in this room is 1.13 m and the frequency is 301 Hz.

Yakvenalex [24]

Answer:

a. 340.13 m/s b. 680.26 m/s c. our wavelength doubles

Explanation:

a. speed of wave, v = fλ were f = frequency = 301 Hz and λ = wavelength = 1.13 m.

v = fλ = 301 Hz × 1.13 m = 340.13 m/s

b. If we double the frequency then f = 2 × 301 Hz = 602 Hz

v = fλ = 602 Hz × 1.13 m = 680.26 m/s

c. If the speed of the wave is still 340.13 m/s, if we cut the frequency in half, then frequency now equals f = 301 Hz/2 = 150.5 Hz.

Since v = fλ,

λ = v/f = 340.13 m/s ÷ 150.5 Hz = 2.26 m.

Since our initial wavelength λ₀ = 1.13 m,

λ/λ₀ = 2.26 m/1.13 m = 2.

So, λ = 2λ₀ our wavelength doubles

5 0

3 years ago