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.
Answer:
Explanation:
The guy wire is making a right angled triangle with the ground and stop sign . It makes an angle of 51 degree with the ground. In this triangle stop sign is the perpendicular and distance from the base on the ground forms the base of the triangle . Wire forms the hypotenuse.
base / hypotenuse = cos51
base = hypotenuse x cos51
= 8 x cos51
= 5.03 ft .
The distance of the stake with which guy wire was attached from the foot of the stop sign is 5.03 ft .
In an experiment if doubling the manipulated variable results in doubling of the responding variable the relationship between the variable is a direct. Having a direct relationship means that the<span> relationship between two numbers or other variables where an increase or decrease in one variable causes the same change to the other variable.</span>
<span>The land between two normal faults moves upward to form a
Answer:D</span><span>
fault-block mountain.</span>
Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):
K = ½mv²
v = √(2K/m)
λ = h/(mv)
= h/(m√(2K/m))
= h/√(2Km)
So λ is proportional to 1/√K.
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:
K = 6U₀
Outer the potential well the potential energy is U₀, so
K = 5U₀
(because kinetic and potential energies add up to 6U₀)
Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is:
1/√(5U₀) : 1/√(6U₀)
= √6 : √5
