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never [62]
4 years ago
8

g Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. Speaker B is 2.00 m to the rig

ht of speaker A. Consider point Q along the extension of the line connecting the speakers, 1.00 m to the right of speaker B. Both speakers emit sound waves that travel directly from the speaker to point Q. (a) What is the lowest frequency for which constructive interference occurs at point Q
Physics
1 answer:
lisabon 2012 [21]4 years ago
7 0

Answer:

172 Hz

Explanation:

We are given that

Distance between speaker A and B=2 m

Distance between speaker B and point Q=1 m

We have to find the lowest frequency for which constructive interference occurs at point Q.

\lambda=\frac{2}{n}

Where n=Integer

When frequency is minimum then wavelength is maximum

when n=1

\lambda_{max}=2

\nu=\frac{v}{\lambda_{max}}

Speed of sound,v=344 m/s

\nu=\frac{344}{2}=172 Hz

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Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial veloc
svet-max [94.6K]

Incomplete question as time is missing.I have assumed some times here.The complete question is here

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Explanation:

Given data

Vi=10 m/s

S=70 m

(a) t₁=0.5 s

(b) t₂=1 s

(c) t₃=1.5 s

(d) t₄=2 s

(e) t₅=2.5 s

To find

Displacement S from t₁ to t₅

Velocity V from t₁ to t₅

Solution

According to kinematic equation of motion and given information conclude that v is given by

v=v_{i}+gt\\

Also get the equation of displacement

S=v_{i}t+(1/2)gt^{2}

These two formula are used to find velocity as well as displacement for time t₁ to t₅

For t₁=0.5 s

v_{1}=v_{i}+gt\\v_{1}=(10m/s)+(9.8m/s^{2} ) (0.5s)\\v_{1}=14.9m/s\\  And\\S_{1} =v_{i}t+(1/2)gt^{2}\\ S_{1}=(10m/s)(0.5s)+(1/2)(9.8m/s^{2} )(0.5s)^{2} \\S_{1}=6.225m

For t₂

v_{2}=v_{i}+gt\\v_{2}=(10m/s)+(9.8m/s^{2} ) (1s)\\v_{2}=19.8m/s\\  And\\S_{2} =v_{i}t+(1/2)gt^{2}\\ S_{2}=(10m/s)(1s)+(1/2)(9.8m/s^{2} )(1s)^{2} \\S_{2}=14.9m

For t₃

v_{3}=v_{i}+gt\\v_{3}=(10m/s)+(9.8m/s^{2} ) (1.5s)\\v_{3}=24.7m/s\\  And\\S_{3} =v_{i}t+(1/2)gt^{2}\\ S_{3}=(10m/s)(1.5s)+(1/2)(9.8m/s^{2} )(1.5s)^{2} \\S_{3}=26.025m

For t₄

v_{4}=v_{i}+gt\\v_{4}=(10m/s)+(9.8m/s^{2} ) (2s)\\v_{4}=29.6m/s\\  And\\S_{4} =v_{i}t+(1/2)gt^{2}\\ S_{4}=(10m/s)(2s)+(1/2)(9.8m/s^{2} )(2s)^{2} \\S_{4}=39.6m

For t₅

v_{5}=v_{i}+gt\\v_{5}=(10m/s)+(9.8m/s^{2} ) (2.5s)\\v_{5}=34.5m/s\\  And\\S_{5} =v_{i}t+(1/2)gt^{2}\\ S_{5}=(10m/s)(2.5s)+(1/2)(9.8m/s^{2} )(2.5s)^{2} \\S_{5}=55.625m

4 0
4 years ago
Which statements describe acceleration? Check all that apply.
DiKsa [7]

Answer:

Negative acceleration occurs when an object speeds up in the negative direction.

Positive acceleration occurs when an object speeds up in the positive direction.

Explanation:

This might be wrong but I get some points for a answer..

8 0
3 years ago
gwendolyn has a mass of 17 kg. she is riding her scooter at 1.15 m/s to the right when she runs into maya. maya has a mass of 13
kykrilka [37]
Inelastic collision happens when two objects joined and move together after the collision

7 0
4 years ago
Ben climbs a tree to retrieve a lost football. If he does 2,400 J of work to climb 8 m, how much force is he using?
krok68 [10]
Work = Force * distance.
d = 8 m
W = 2400 J
F = ???

2400 = 8 * F
F = 2400/8
F = 300 N


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4 years ago
Pen and ink manufacturers are asked to submit their new ink formulations to what database?
Ilia_Sergeevich [38]
The international ink library.
3 0
3 years ago
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