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german
3 years ago
13

1. In order to change the speed of a wave, you must change something about the medium. T or F

Physics
1 answer:
Slav-nsk [51]3 years ago
7 0

Answer:

1) False

As the frequency can be defined as the number of times wave formed per unit time or in other words the numbers of a combination of a crest and trough forms. So the frequncy of a sound wave depends upon the source producing sounds

2)False

Diffraction is the process in which a light wave bends when they pass through a specified medium such as water.

3)True

hope it helps

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A light wave passes through an aperture (that is, a narrow slit). When it does so, the degree to which the wave spreads out will
crimeas [40]

Explanation:

Single slit diffraction

Diffraction is the phenomenon of spreading out of waves as they pass through an aperture or around objects. Diffraction occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave. For very small aperture sizes, the vast majority of the wave is blocked. in case of  large apertures the wave passes by or through the obstacle without any significant diffraction.

7 0
3 years ago
A force of 9 pounds stretches a spring 1 foot. A mass weighing 6.4 pounds is attached to the spring, and the system is then imme
mart [117]

Answer:

\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0

Explanation:

let m be the mass attached, let k be the spring constant and let \beta be the positive damping constant.

-By Newton's second law:

m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}

where x(t) is the displacement from equilibrium position. The equation can be transformed into:

\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0  shich is the equation of motion.

7 0
3 years ago
A skater is using very low friction rollerblades. A friend throws a Frisbee at her, on the straight line along which she is coas
kupik [55]

Answer:

a)  perfectly inelastic,  b)  collision is inelastic,  c)   elastic  

Explanation:

In this exercise, it is asked to identify what type of shock occurs between the skater and the frisbee, for this we must define a system formed by the skater and the fribee, so that the forces during the crash have been internal and the amount of movement is preserved

Initial instant. Before the skater touches the frisbee

    p₀ = M v₁ + m v₂

where M and m are the masses of the skater and frisbee, respectively

for the final moment they give us several possibilities, in all case the moment is conserved

       p₀ = p_{f}

case a)

Final instant. grabs the frisbee and holds it

    p_{f} = (M + m) v '

     p₀ = p_{f}

We can see that this shock is perfectly inelastic, it holds the fressbee

case b)

final instant.

This case is similar to the previous one, but the final speed of fresbee is zero, therefore this collision is inelastic and the kinetic energy is not conserved.

case c)

final instant. Grab the fressbee and resend it

      p_{f} = M v_{1f} + m v_{2f}

this is an elastic Shock since the equivalent of a rebound of the fressbee, the kinetic energy is conserved.

5 0
3 years ago
A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanba
Phantasy [73]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

\mu = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

m_1 = Mass of bean bag = 0.354 kg

m_2 = Mass of empty crate = 3.77 kg

v_1 = Speed of the bean bag

v_2 = Speed of the crate

Acceleration

a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g

a=--9.81\times 0.48=4.7088\ m/s^2

From equation of motion

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s

In this system the momentum is conserved

m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s

The speed of the bean bag is 31.42383 m/s

8 0
3 years ago
Find the emitted power per square meter of peak intensity for a 3000 k object that emits thermal radiation.
Aleksandr [31]
According to Stefan-Boltzmann Law, the thermal energy radiated by a radiator per second per unit area is proportional to the fourth power of the absolute temperature. It is given by;
    P/A = σ T⁴ j/m²s 
Where; P is the power, A is the area in square Meters, T is temperature in kelvin and σ is the Stefan-Boltzmann constant, ( 5.67 × 10^-8 watt/m²K⁴)
Therefore;
Power/square meter = (5.67 × 10^-8) × (3000)⁴
                                 =  4.59 × 10^6 Watts/square meter
8 0
3 years ago
Read 2 more answers
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