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3 years ago

Which wave diffracts the most when encountering an obstacle?

Physics

1 answer:

viva [34]3 years ago

8 0

Answer;

-A wave with the longest wavelength.

Explanation;

-Diffraction is the apparent of wave through,around small obstacles and the spreading out of wave past small openings. When thinking of diffraction of a wave think of shining a flashlight around a corner. The light bends around the corner but there is a place where it is dark and the light does not hit. Diffraction of a wave is basically the wave bending around an object then dispersing out.

-The amount of diffraction (the sharpness of the bending) increases with increasing wavelength and decreases with decreasing wavelength. When the wavelength of the waves is smaller than the obstacle, no noticeable diffraction occurs.

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Imagine you could travel to the moon where the acceleration due to gravity is 1.6 m/s^2. What would be the period of

VladimirAG [237]

Answer:

4.9612 s

Explanation:

Applying,

T = 2π√(L/g)............... Equation 1

Where T = period of the pendulum, L = Lenght of the pendulum, g = acceleration due to gravity of the moon, π = pie.

From the question,

Given: L = 1 m, g = 1.6 m/s²

Constant: π = 3.14

Substitute these values into equation 1

T = 2×3.14×√(1/1.6)

T = 6.28√(0.625)

T = 6.28×0.79

T = 4.9612 s

6 0

3 years ago

1. A car with a mass of 2500 kg accelerates when the traffic light turns green. If the net force

soldier1979 [14.2K]

;Net force = mass of the body × acceleration of the body due to the net force

; 5000 = 2500 a...then divide both sides by 2500

; acceleration(a) = 2 m/s^2

5 0

3 years ago

A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl

Alex

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

$Q = C_w BH^{3/2}$

$H = (\dfrac{Q}{C_wB})^{2/3}$

$H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}$

$H = 0.371\ m$

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

7 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago

The equation for momentum is p=mv. Which letter represents the momentum? Question 2 options: p m v

vovikov84 [41]

P stands for momentum

8 0

3 years ago