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3 years ago

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Resonance occurs when an object vibrating at or near the resonant frequency of a second object to vibrate. What form of waves ar

e involved in resonance

1 answer:

Ivanshal [37]3 years ago

8 0

Answer:

Resonance depends on objects, this may happen for example when you play guitar in a given room, you may find that for some notes the walls or some object vibrate more than for others. This is because those notes are near the frequency of resonance of the walls.

So waves involved are waves that can move or affect objects (in this case the pressure waves of the sound, and the waves that are moving the wall).

this means that the waves are mechanic waves.

Now, in electromagnetics, you also can find resonance frequencies for electromagnetic waves trapped in things called cavities, but this is a different topic.

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The main difference between internal combustion engine and the diesel engine is?

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<span>The difference between a internal combustion engine and a diesel engine is the ignition, But a Diesel engine is an internal combustion engine. The both burn internal one uses compression to fire the other uses ignition system.</span>

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3 years ago

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A gecko crawls vertically up and down a wall. Its motion is shown on the following graph of vertical position yyy vs. time ttt.

earnstyle [38]

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3 years ago

A 2 L balloon filled with gas is warmed from 280 K to 700 K. What is the volume of the gas after it is heated?

irakobra [83]

Answer:

New volume, v2 = 0.8L

Explanation:

<u>Given the following data;</u>

Original Volume = 2L

Original Temperature = 280K

New Temperature = 700K

To find new volume V2, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

$VT = K$

$\frac{V1}{T1} = \frac{V2}{T2}$

$\frac{V1}{T1} = \frac{V2}{T2}$

Making V2 as the subject formula, we have;

$V_{2}= \frac{V1}{T1} * T_{2}$

$V_{2}= \frac{2}{700} * 280$

$V_{2}= 0.0029 * 280$

V2 = 0.8L

Therefore, the volume of the gas after it is heated is 0.8L.

7 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

3 years ago

To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle wi

Gnom [1K]

Answer:

(a) Acceleration of positron is 6.03 x 10¹³ m/s²

(b) Speed of positron after 8.70 x 10⁻⁹ s is 5.24 x 10⁵ m/s

Explanation:

Given :

Constant electric field, E = 343 N/C

Mass of positron, m = 9.1 x 10⁻³¹ kg

Charge of positron, q = +e = 1.6 x 10⁻¹⁹ C

(a) Coulomb force on the positron is determine by the relation :

F = q x E ....(1)

But, force is also equals to product of mass and acceleration. So,

F = ma .....(2)

Here a is acceleration.

From equation (1) and (2).

m x a = q x E

$a=\frac{qE}{m}$

Substitute the values of q, E and m in the above equation.

$a=\frac{1.6\times10^{-19}\times 343}{9.1\times10^{-31} }$

a = 6.03 x 10¹³ m/s²

(b) Initially, the positron is at rest, so its initial speed is zero.

The equation of motion for positron is :

v = u + at

Here v is final speed, u is initial speed and t is time.

Since, u is zero, so the equation becomes :

v = at

Substitute 8.70 x 10⁻⁹ s for t and 6.03 x 10¹³ m/s² for a in the above equation.

v = 6.03 x 10¹³ x 8.70 x 10⁻⁹ m/s

v = 5.24 x 10⁵ m/s

6 0

3 years ago