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1 year ago

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A sound technician is testing the sound acoustics in a theatre for an upcoming music concert. He sets up speakers in different l

ocations and tests the sound levels in various seats in the theater. In one seat, the sound is twice as loud as it is in other seats in the theater. What wave phenomenon best explains why the sound is louder in one seat compared to others? (1 point)
Reflection off the theater walls

Diffraction between speakers

Constructive interference

Destructive interference

1 answer:

velikii [3]1 year ago

4 0

Consertive alex dggghh

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A 500 kg motorcycle accelerates at a rate of 2 m/s .how much force was applied to the motorcycle?

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The graph below shows the sunspot number observed between 1750 and 2000. The graph shows sunspot number on the y axis and years

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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The magnets need to be placed with red closest to blue.

Opposite poles attract.

The magnets will be attracted to each other with enough force to stick together.

Explanation:

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A 3.0-kg cart moving to the right with a speed of 1.0 m/s has a head-on collision with a 5.0-kg cart that is initially moving to

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Answer:

$v = 0.8 m/s$ towards left

Explanation:

As we know that there is no external force on the system of two cart so total momentum of the system is conserved

so we will say

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now plug in all data into the above equation

$3(1) + 5(-2) = 3(-1) + 5 v$

here we assumed that left direction of motion is negative while right direction is positive

so we can solve it for speed v now

$3 - 10 = - 3 + 5 v$

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$v = -0.8 m/s$

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