Help me with these questions PLZ

Two speakers separated by a distance of 4.40 m emit sound. The speakers have opposite phase. A person listens from a location 3.

Hoochie [10]

Answer:

f = 147.21 Hz

Explanation:

In order to have a destructive interference, as the source emit in opposite phases, the path difference between the distance to the person, measured in a straight line from the speakers, must be equal to an integer number of wavelengths.

We need to know the distance from the listener to the other speaker, located 4.4 m from the one which is directly in front of him, which we can find using Pythagorean theorem, as follows:

l₂ = √(3)²+(4.4)² = 5.33 m

The difference in path will be, then:

d = l₂-l₁ = 5.33 m - 3.00 m = 2.33 m

For the lowest frequency that gives destructive interference, the wavelength will be highest possible, which happens when the distance is just one wavelength.

⇒ d = λ = 2.33 m

In any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

v = λ*f Κ ⇒ f = v/λ

Taking the speed of sound as 343 m/s, and solving for f, we get:

f= 343 m/s / 2.33 m = 147.21 Hz

3 0

3 years ago

A skateboarder shoots off a ramp with a velocity of 5.1 m/s, directed at an angle of 55° above the horizontal. The end of the ra

ExtremeBDS [4]

Answer

given,

initial velocity of skateboard = 5.1 m/s

angle above the horizontal = 55°

height of the ramp = 1 m

a) maximum height of projectile

$H = \dfrac{u^2sin^2 \theta}{2g}$

$H = \dfrac{5.1^2\times sin^2 55^0}{2\times 9.81}$

H = 0.889 m

the maximum height of the skateboard above the ground

= 1 + 0.889

= 1.889 m

b) time to reach the height

$t = \dfrac{u\ sin\theta}{g}$

$t = \dfrac{5.1\ sin55^0}{9.8}$

t = 0.426 s

horizontal distance = u cos θ × t

= 5.1 × cos 55° × 0.426

horizontal distance = 1.25 m

7 0

3 years ago

3. why is the sum of the maximum voltages across each element in a series r l c circuit usually greater than the maximum applied

Llana [10]

The sum of the maximum voltages across each element in a series RLC circuit is usually greater than the maximum applied voltage because voltages are added by vector addition.

<h3>What is the Kichoff's loop rule?</h3>

Kirchhoff's loop rule states that the algebraic sum of potential differences, as well as the voltage supplied by the voltage sources and resistances, in any loop must be equal to zero.

In a series RLCcircuit, the voltages are not added by scalar addition but by vector addition.

Kirchhoff's loop rule is not violated since the voltages across different elements in the circuit are not at their maximum values.

Therefore, the sum of the maximum voltages across each element in a series RLC circuit is usually greater than the maximum applied voltage because voltages are added by vector addition.

Learn more about Kichoff's loop rule at: https://brainly.in/question/35360816

#SPJ1

8 0

2 years ago

A zero-order reaction has a constant rate of 2.30×10−4 M/s. If after 80.0 seconds the concentration has dropped to 1.50×10−2 M,

dolphi86 [110]

Answer:

Initial concentration of the reactant = 3.34 × 10^(-2)M

Explanation:

Rate of reaction = 2.30×10−4 M/s,

Time of reaction = 80s

Final concentration = 1.50×10−2 M

Initial concentration = Rate of reaction × Time of reaction + Final concentration

= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M

Initial concentration = 3.34 × 10^(-2)M

6 0

2 years ago

provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small

gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

$T_{simple$ = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / $I$ ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and $I$ is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

$I$ = $\frac{1}{3}$ mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/ $I$ ) OR T = 2π√( $I$ /mgL)

so we can use $I$ = $\frac{1}{3}$ mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L = $\frac{1}{2}$ D.

now, substituting these equations, the period becomes;

T = 2π/√( $I$ /mgL) OR T = $2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }$ OR T = 2π√(2D/3g ) ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω $_{simple$ = 2π/ $T_{simple$ OR ω $_{simple$ = √(g/D) OR ω $_{simple$ = 2π√( D/g )

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × $T_{simple$

we substitute in value of $T_{simple$

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

8 0

2 years ago