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Rama09 [41]
3 years ago
12

If you are driving an oscillatory system at a certain frequency, but the amplitude is much smaller than it could be, you can be

certain that
1. The driving frequency is too low.
2. The driving frequency is too high.
3. The driving frequency is not matched to the natural frequency of the oscillatory system.
Physics
1 answer:
kakasveta [241]3 years ago
5 0

Answer:

3. The driving frequency is not matched to the natural frequency of the oscillatory system.

Explanation:

  • It is the condition of resonance when the one body that oscillates with a maximum amplitude when the frequency of the applied force is equal to the natural frequency of the body.
  • Every body has its own natural frequency.
  • Here the driving force may be more or less but we are sure that it is not equal to the natural frequency of oscillatory system. Hence the force is not in resonance with the oscillatory system.

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Katen [24]
Within that unit of blood, you are going to have plasma, white blood cells, red blood cells, and platelets. 
7 0
3 years ago
Identify each of the quantities below as either vector or scalar
natta225 [31]

Answer:

2nd answer 3.2m/s is a scalar l, all others are vectors.

Explanation:

In order to be considered as a vector a certain quantity must have both a magnitude and a direction.

The most common example is velocity. It has both magnitude and direction, like the 1st and the last answers.

The 3rd answer is an acceleration. It is also given with its direction. So we have to consider it also as a vector.

But 2nd answer has no direction mentioned, so it is considered as a scalar.

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3 years ago
Help me with the following problem
Snezhnost [94]

The electric field at arbitrary point outside the sphere is determined as the E = σr³/k.

<h3>Electric field determined from Gauss law</h3>

The electric field of the surface is determined from Gauss law as shown below;

E ∫ds = Q/ε

E (4πr²) = Q/ε

E = Q/4πεr² . r

E = \frac{Q R}{4\pi \varepsilon r^3}

<h3>Electric field outside the sphere with dielectric with polarization</h3>

P = \frac{\sigma}{E} \\\\E = \frac{\sigma}{P}

E = \frac{\sigma}{k/r^3} \\\\E = \frac{\sigma r^3}{k}

where;

  • σ is dipole moment of atom of the metal
  • k is dielectric constant

Thus, the electric field at arbitrary point outside the sphere is determined as the E = σr³/k.

The complete question:

A metal sphere of radius R carries a total charge Q. outside the sphere is a dielectric with polarization p(f) k/r^3er. Determine the electric field at arbitrary point outside the sphere.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

6 0
2 years ago
The 30-kg disk is originally spinning at v = 125 rad&gt;s. If it is placed on the ground, for which the coefficient of kinetic f
irga5000 [103]

Answer:

t = 3.82 s

Ax = 147 N  (←)

Ay = 294 N   (+↑)

Explanation:

Given

m = 30.0 Kg

ωinitial = 125 rad/s

ωfinal = 0 rad/s

μC = 0.5

R = 0.3 m

t = ?

Ax = ?

Ay = ?

For the disk, we can apply

∑ τ = I*α

where I = m*R²/2

then

⇒ R*Ffriction = (m*R²/2)*α

⇒ R*(-μC*N) = R*(-μC*m*g) = (m*R²/2)*α

⇒ α = -2*μC*g / R

⇒ α = -2*(0.5)*(9.8) / 0.3 = -32.666 rad/s²

we can use the equation to get t:

α = Δω / t      ⇒   t = Δω / α = (0 - 125) / (-32.666)

⇒   t = 3.82 s

The horizontal and vertical components of force which the member AB exerts on the pin at A during this time are

∑ Fx = 0  (+→)

Ax - Ffriction = 0

⇒  Ax = Ffriction = μC*m*g = (0.5)*(30)*(9.8) = 147 N

⇒   Ax = 147 N  (←)

∑ Fy = 0   (+↑)

⇒  Ay - m*g = 0

⇒  Ay = m*g

⇒  Ay = 30*9.8 = 294 N

⇒  Ay = 294 N   (+↑)

5 0
3 years ago
In a light wave, what properties tell you the color of light
sesenic [268]

Answer:

Answer :- In a light wave the property of wave which tells about the color of light is it's Wavelength .

Wavelength is the distance between one crest and one through , also it is the distance after which the wave repeat itself !

It's SI unit is meter !

It is scalar quantity !!

Different Wavelength of light have different color !!

• VIBGYOR

i.e, Violent , Indigo , Blue , Green , Yellow Orange, and Red along with their shades are the colors which we can see !!

• They almost range from 400nm to 700nm ( visible range of light ) !!

4 0
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