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3 years ago

7

A speaker that emits sound is located at the origin of a coordinate system. Two microphones are located on the x-axis, with micr

ophone 1 at the x = 3.0 meter mark and microphone 2 at the x = 5.0 meter mark. If the speed of sound is 343 m/s, how much longer does it take a sound emitted by the speaker to reach microphone 2 than microphone 1?

1 answer:

Taya2010 [7]3 years ago

5 0

Answer:

0.00583 seconds

Explanation:

Distance from mic 1 to origin = 3 m

Distance from mic 2 to origin = 5 m

Speed of sound = 343 m/s

Time taken by mic 1

$t_1=\frac{\text{Distance from mic 1 to origin}}{\text{Speed of sound}}\\\Rightarrow t_1=\frac{3}{343}=0.008746\ s$

$t_2=\frac{\text{Distance from mic 2 to origin}}{\text{Speed of sound}}\\\Rightarrow t_2=\frac{5}{343}=0.014577\ s$

Time difference = t₂ - t₁ = 0.14577-0.008746 = 0.00583 s

∴ Difference in time taken by the speaker is 0.00583 s

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3 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

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Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

$0=F-FrictionAB$

$F=FrictionAB=Nab*$ μs

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$0=Nab-W$

$Nab=W=mg$

replacing this we have:

$F=$ μs $*Nab=$ μs* $mg=41.6N$

and μs $=41.6N/(mg)$

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

$F-Friction1(ground)-Friction2(AB)=0$

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$F=Friction1(ground)+Friction2(AB)$

we know Friction2(AB) because we found that in the previous block so:

$F=Friction1(ground)+mg$ *μs

for the other friction we have to use the equation:

$Friction(ground)=N(ground)$ *μs

from y axis we have:

$N(ground)-w-Normal(AB)=0$

$N(ground)=w+Normal(AB)$

we found the value of Normal(AB) with the previous block so:

$N(ground)=mg+mg=2mg$

and:

$Friction(ground)=2mg$ *μs

$F=Friction(ground)+mg$ *μs

$F=2mg$ *μs+μs* $mg=3mg$ *μs

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3 years ago

A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

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Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

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Kinetic friction = 0.4

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Using formula of friction force

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Put the value into the formula

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So, the tension must exceeds 29.6 N for the block to move

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Using formula of frictional force

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Put the value in to the formula

$f=0.4\times37$

$f=14.8\ N$

Hence, The frictional force acting on the block is 14.8 N.

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