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Ray Of Light [21]

2 years ago

12

Which of these systems is an oscillator? A. A barrel rolling down the hill B. A child sitting on a swing a skater falling on the

ice C. A skater falling on the ice D. A car going uphill

1 answer:

Paladinen [302]2 years ago

7 0

Answer:

Option B:

A child sitting on a swing.

Explanation:

When we hear the word oscillator, a good example is the pendulum bob of a grandfather clock. We can picture the motion to get a perfect understanding of its path of motion and relate it to other systems of motion in our everyday life.

An oscillator is a system that moves in such a way that it reverses its direction after a period of time. It can be seen as a "to-and-fro" motion.

From the options, a child sitting on a swing is the perfect example of an oscillating system because the child will be moving forwards and backwards, alternately reversing the direction of motion with time.

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The bandgap of InP semiconductor laser is 1.0 eV. The effective mass of the valence band is ½ of the effective mass of the condu

klemol [59]

Answer: the wavelength of this transition is 1.2039 um

Explanation:

Given that;

the energy level between the transitioning energy gap Eg = 1.0 + 0.03 = 1.03 eV

we know that λ = 1.24 / Eg

so we substitute our Eg into the above equation

λ = 1.24 / 1.03

λ = 1.2039 um

therefore the wavelength of this transition is 1.2039 um

4 0

3 years ago

It is easier to roll a stone up a sloping road than to lift it vertical upwards because.

kvasek [131]

Answer:

Because the force one needs to apply is reduced.

Explanation:

Work can be expressed in terms of force and distance as follows,

Work = Force × distance

To be more precise, the selection of these two parameters should be in a manner that the distance is what the line of action of the force traces.
It is obvious that for a given amount of work when one parameter is changed the other changes to counteract the change of the first one.
Imagine the situation shown in the attached figure. The amount of work that needs to be done during the lifting of the mass to a height of $\small h$ vertically is $E = mg\times h = mgh$ .
On the slope, however, one needs to push the object an $L$ distance up to get the object to that $h$ height ultimately.
The work he does on the slope is $mgh = F\times L$ and since $L > h$ , $F < mg$ .
The force that needs to be applied is less than that applied in the vertical lift. Hence it feels easier. #SPJ4

3 0

1 year ago

Which of the following statements are true concerning electromagnetic radiation fields?Select all that apply.a.) The electric an

ale4655 [162]

Answer:

a.) The electric and magnetic fields are in phase with each other as they propagate through space.

Explanation:

Electromagnetic wave is a transverse wave in which magnetic field and electric field both induces each other as both changes with time

Here magnetic field induces electric field and similarly magnetic field induces electric field.

As we know that this is a transverse wave so here magnetic field and electric field lies in perpendicular planes. but they both propagate in same direction in such a wave that both fields reaches their maximum position and minimum positions simultaneously

So the correct answer is

a.) The electric and magnetic fields are in phase with each other as they propagate through space.

6 0

3 years ago

A 0.150 kg mass is attached to a spring with k = 18.9 N/m. At the equilibrium position, it moves 2.39 m/s. How much mechanical e

Nastasia [14]

Answer:0.428

Explanation: got it right on Acellus

5 0

3 years ago

On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass

Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

$\theta=53^{\circ}$

$\mu_k=0.16$

Speed,v=1.5m/s

Net force in x direction must be zero

$F_{net}=0$

$Fsin\theta-f=0$

$Fsin\theta=f$

Net force in y direction

$N-mg-Fcos\theta=0$

$N=mg+Fcos\theta$

$f=\mu_kN=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_kmg+\mu_kFcos\theta$

$Fsin\theta-\mu_kFcos\theta=\mu_kmg$

$F(sin\theta-\mu_kcos\theta)=\mu_kmg$

$F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}$

Power,P=Fv

$P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v$

Where $g=9.8m/s^2$

B.Substitute the values

$P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5$

$P=40.17W$

6 0

3 years ago