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3 years ago

Why do you think fixed boundaries ""flip"" waves and loose boundaries do not?

Physics

1 answer:

Vinil7 [7]3 years ago

5 0

Answer:

When the obstacle is fixed, the law of action and reaction, makes the reflected wave is inverted.

When the obstacle is mobile, he mobile point, it moves in the direction of the wave, therefore there is no inversion of it.

Explanation:

Waves when they reach an obstacle behave like a shock, therefore if we use the conservation of momentum the wave must reverse its speed, this explains that the speed changes sign, the wave is reflected.

When the obstacle is fixed, the wave when it reaches the obstacle exerts a force on the point, by the law of action and reaction the point exerts on the wave a force of equal magnitude but in the opposite direction, this reaction force which makes the reflected wave is inverted.

When the obstacle is mobile, this is without friction, when the wave arrives it exerts a force on the mobile point, it moves in the direction of the wave, reaching the maximum amplitude of the incident wave, when it is reflected the point begins to go down along with the wave, therefore there is no inversion of it.

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VMariaS [17]

Answer:

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4 years ago

A scientist heated a tank containing 50 g of water. The specific heat of water is 4.18 J/gºC. The temperature of the water incre

Amanda [17]

I got 5,225 by 50x4.18= 209(25)=5,225

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3 years ago

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0

3 years ago

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0

3 years ago

He does whatever a spider can. Spider-Man, who has a mass of 76 [kg], is clinging onto an inclined wall forming an inclination a

Aliun [14]

Answer:570.54 N

Explanation:

Given

mass of man=76 kg

$\theta =50^{\circ}$

As man is standing over inclined building therefore

its weight has two components i.e. sin and cos component

Force perpendicular to inclined wall

$F=mgcos\theta =76\times 9.8\times \sin 50$

F=570.54 N

4 0

3 years ago