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trapecia [35]
3 years ago
11

Speakers A and B are vibrating in phase. They are directly facing each other, are 6.69 m apart, and are each playing a 75.0-Hz t

one. The speed of sound is 343 m/s. What is the distance from speaker A to the first point on the line between the speakers where constructive interference occurs?
Physics
1 answer:
maw [93]3 years ago
3 0

Answer:

3.117 m

Explanation:

Given that:

the distance of separation between speaker A and speaker B (L) = 6.69 m

Frequency (F) = 750 -Hz tone

Velocity of speed of sound = 343 m/s

The distance from Speaker A to the first point (L₁) on the line can be calculated by using the formula:

L_1=\frac{L-A}{2}

where A = \frac{Velocity ofthe sound (V)}{Frequency (F)}

we have:

L_1=\frac{L-\frac{V}{F} }{2}

L_1=\frac{6.69-\frac{343}{750} }{2}

L_1=\frac{6.69-0.457 }{2}

L_1=\frac{6.233 }{2}

L_1= 3.1165 m

L_1=3.117 m

∴ the distance from speaker A to the first point on the line between the speakers where constructive interference occurs = 3.117 m

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Softa [21]

Answer:

the work that must be done to stop the hoop is 2.662 J

Explanation:

Given;

mass of the hoop, m = 110 kg

speed of the center mass, v = 0.22 m/s

The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;

W = ΔK.E

W = ¹/₂mv²

W = ¹/₂ x 110 x 0.22²

W = 2.662 J

Therefore, the work that must be done to stop the hoop is 2.662 J

6 0
3 years ago
Photographs of many young stars show long jets of material apparently being ejected from their poles.
Rudiy27

Answer:

A) True

Explanation:

Researchers have detected numerous jets of gas ejected from poles of young stars and planetary nebulae.

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7 0
3 years ago
what us the velocity in meters per second of a runner who runs exactly 110 m toward the beach in 72 secs
finlep [7]

Velocity = Distance / time taken

   =  110 m /72s

     1.52  ms⁻¹

4 0
3 years ago
Two charges that are separated by one meter exert 1-n forces on each other. if the magnitude of each charge is doubled, the forc
Goshia [24]
The electrostatic force between the two charges is
F=k_E  \frac{q_1 q_2}{r^2}
where q1 and q2 are the magnitudes of the two charges, and r the distance between them.

We can see from the formula that F is proportional to the product between the two charges:
F \sim q_1 q_2
so, if the magnitude of each charge is doubled, the new force will get a factor 4:
F' \sim (2 q_1 )(2 q_2 )=4 q_1 q_2 =4 F
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5 0
3 years ago
In a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section and a
timofeeve [1]

Answer:

a) F₁ = 1.48 x 10³ N

b) P = 1.88*10⁵ Pa

c) The work  is equal in both pistons

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F₁) on a small area piston (A₁), then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F₂) can be exerted that is proportional to the area(A₂) of the piston.

Pressure is defined as the force per unit area:

P=\frac{F}{A}  Formula (1)

P₁=P₂

\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} } Formula (2)

Data

r₁= 5 cm = 0.05 m

r₂= 15 cm = 0.15 m

F₂=  13300N

Area of the pistons (A₁,A₂)

A=π*r² : Area of the circle

A₁ = π*(0.05)²=7.85*10⁻³ m²

A₂= π*(0.15)²= 70.69*10⁻³ m²

a) Force that compressed air must exert to lift a car weighing 13300 N

We replace data in the formula (2)

\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} }

F_{1} = \frac{13300*7.85*10^{-3} }{70.69*10^{-3} }

F₁ =  1.48 x 10³ N

b) Air pressure produced by F₁

We replace data in the formula (1)

P=\frac{F}{A}

F₁ =  1.48 x 10³ N , A₁ = 7.85*10⁻³ m²

P=\frac{1.48*10^{3} }{7.85*10^{-3} }

P= 1.88*10⁵ Pa

c)The volume of liquid displaced by the small piston is distributed in a thin layer on the large piston, so that the product of the force by the displacement (the work) is equal in both pistons.

3 0
3 years ago
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