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3 years ago

12

The two loudspeakers in the drawing are producing identical sound waves. The waves spread out and overlap at the point P. What i

s the difference in the two path lengths if point P is at the third sound intensity minimum from the central sound intensity maximum

1 answer:

Damm [24]3 years ago

5 0

Answer:

5/2π

Explanation:

According to quizlet the answer is 5/2π

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A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3 km/h.

Amanda [17]

Answer:

$F_{1}=\frac{1}{5}F_{2}$ or $F_{2}=5F_{1}$

In other words, $F_{1}$ is one fifth of $F_{2}$ or $F_{2}$ is five times as big as $F_{1}$

Explanation:

In order to solve this problem we must start by sketching the situation (refer to the attached picture).

When the ship is pulled only by force 1, it will change its speed by 3km/hr in 10 seconds. So in order to use these values we need to either turn the km/hr in km/s or turn the seconds to hours. Let's turn the seconds to hours:

$10s*\frac{1hr}{3600s}=\frac{1}{360} hr$

so we can now use the acceleration formula to find the acceleration of the boat so we get:

$a=\frac{\Delta v}{\Delta t}$

which will give us an accceleration of:

$a=\frac{3km/hr}{\frac{1}{360}hr}=1080km/hr^{2}$

once we got the acceleration we can for sure say taht:

$F_{1}=ma=m*1080\frac{km}{hr^{2}}$

Now, if we take a look at the second drawing we can see that the resultant force applied to the boat is found by adding the two forces, force one and force two, so we get:

$F_{1}+F_{2}=ma$

in this case the acceleration changes because the change in velocity is of 18km/hr in the same 10 seconds, so we get that:

$a=\frac{\Delta v}{\Delta t}$

$a=\frac{18km/hr}{\frac{1}{360}hr}=6480km/hr^{2}$

so we can say that:

$F_{1}+F_{2}=m*6480km/hr^{2}$

we can substitute the first force into this equation so we get:

$m*1080km/hr^{2}+F_{2}=m*6480km/hr^{2}$

and solve for the second force, so we get:

$F_{2}=m*6480km/hr^{2}-m*1080km/hr^{2}$

which yields:

$F_{2}=m*5400km/hr^{2}$

Now we can compare theh two forces, force 1 and force 2 by dividing them:

$\frac{F_{1}}{F_{2}}=\frac{m*1080km/hr^{2}}{m*5400km/hr^{2}}$

which yields:

$\frac{F_{1}}{F_{2}}=\frac{1}{5}$

when solving for the first force we get:

$F_{1}=\frac{1}{5}F_{2}$

which tells us that the second force is one fifth of the first force.

and when solving for the second force we get that:

$F_{2}=5F_{1}$

which means that the second force is 5 times as big as the first force.

8 0

3 years ago

The aorta is approximately 25 mm in diameter. The mean pressure there is about 100 mmHg and the blood flows through the aorta at

Ksenya-84 [330]

Answer:

Explanation:

25 mm diameter

r₁ = 12.5 x 10⁻³ m radius.

cross sectional area = a₁

Pressure P₁ = 100 x 10⁻³ x 13.6 x 9.8 Pa

a )

velocity of blood v₁ = .6 m /s

Cross sectional area at blockade = 3/4 a₁

Velocity at blockade area = v₂

As liquid is in-compressible

a₁v₁ = a₂v₂

a₁ x .6 m /s = 3/4 a₁ v₂

v₂ = .8m/s

b )

Applying Bernauli's theorem formula

P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²

100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ + 1/2x 1060 x .8²

13328 +190.8 = P₂ + 339.2

P₂ = 13179.6 Pa

= 13179 / 13.6 x 10³ x 9.8 m of Hg

P₂ = .09888 m of Hg

98.88 mm of Hg

8 0

3 years ago

What were the patterns that people saw in the sky long ago?

Lena [83]

Long time ago, people saw the constellations as patterns in the sky. They names these patterns and tell stories about them. What people saw laong time ago are just mere patterns which forms animals and shapes. We got the names of our constellations from the Greeks who named the constellations after the mythological heroes and mythological legends.

7 0

3 years ago

Read 2 more answers

A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this

mars1129 [50]

Answer:

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

a.

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>

<u></u>

b.

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

<u>W = 810.6 J</u>

7 0

3 years ago

What happens to the electrostatic force between two charged particles if the distance between them is doubled?

Stells [14]

According to Coulomb's Law , The size of the force varies inversely as the square of the distance between the two charges. So ,if the distance between the two charges is doubled, the electrostatic force will become weak by one fourth of the original force.

5 0

3 years ago