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Leto [7]
3 years ago
13

Two speakers face each other, and they each emit a sound of wavelength λ. One speaker is 180∘ out of phase with respect to the o

ther. If we separate the speakers by a distance 1.5λ, how far from the left-most speaker should we place a microphone in order to pick up the loudest sound? Ignore reflections from nearby surfaces. Select all that apply.. .
a) 1/4λ.
b) λ. c
) 0λ.
d) 1/2 λ. e) 3/4 λ.
Physics
1 answer:
kolbaska11 [484]3 years ago
4 0
The question ask to calculate and choose among the following is the distance of the microphone from the left - most speaker  in order to pick up the loudest  sound and the best answer would be letter C. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications 
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A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a
GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}

T = 7.83 X10⁻⁷ s

6 0
2 years ago
A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i
snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle=45^{\circ}

Acceleration due to gravity on earth is 9.8 m/s^2

Acceleration due to gravity on moon is \frac{9.8}{6}=1.63 m/s^2

Range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

R_{earth}=\frac{u^2\sin 2\theta }{g}=5----1

R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}-----2

Divide 1 & 2

\frac{5}{R_{moon}}=\frac{1}{6}

R_{moon}=30 m

4 0
3 years ago
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