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nordsb [41]
3 years ago
6

Explain, using your own words, how noise cancelling headphones work, (physics)

Physics
1 answer:
Afina-wow [57]3 years ago
4 0

Answer:

They use noise control, creating a wave that negates outside or ambient sound and replaces it with the desired sound that listeners request.

Explanation:

I hope this helped

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You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in
FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, V=108\ cm^3

We know that the volume of the cylinder is given by :

V=\pi r^2 h

108=\pi r^2 h    

h=\dfrac{108}{\pi r^2}............(1)

The upper area is given by :

A=32r^2+2\pi rh

A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}

A=32r^2+\dfrac{216}{r}

For maximum area, differentiate above equation wrt r such that, we get :

\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}

64r-\dfrac{216}{r^2}=0

r^3=\dfrac{216}{64}

r = 1.83 m

Dividing equation (1) with r such that,

\dfrac{h}{r}=\dfrac{108}{\pi r}

\dfrac{h}{r}=\dfrac{108}{\pi 1.83}

\dfrac{h}{r}=59 \pi

Hence, this is the required solution.

8 0
3 years ago
__________ are scientists that study the composition of earth including rock layers at the surface.
insens350 [35]

Answer:

Geology is the study of planet Earth, including its composition and structure. Geologists are scientists who study Earth and the processes that shape Earth over time. Geologists study two types of forces that change Earth's surface.

Explanation:

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4 0
2 years ago
A person wants to fire a water balloon cannon such that it hits a target 100m100m away. if the cannon can only be launched at 45
vladimir2022 [97]
<span>31.3 m/s Since the water balloon is being launched at a 45 degree angle, the horizontal and vertical speeds will be identical. Also the time the balloon takes to reach its peak altitude will match the time it takes to fall. So let's create a few expressions about what we know. Distance the water balloon travels at velocity v for time t d = vt Total time required for the entire trip is double since the balloon goes up, then goes down t = 2v/a Now let's plug in the numbers we have, assuming the acceleration due to gravity is 9.8 m/s^2 t = 2v/9.8 100 = vt Substitute 2v/9.8 for t in the 2nd formula 100 = v(2v/9.8) Solve for v. 100 = v(2v/9.8) 100 = 2v^2/9.8 980. = 2v^2 490 = v^2 22.13594 = v So we now know that both the horizontal velocity and vertical velocity needed is 22.13594 m/s. Let's verify that 2*22.13594 / 9.8 = 4.51754 So it will take 4.51754 second for the balloon to hit the ground after being launched. 4.51754 * 22.13594 = 100 And during that time it will travel 100 meters horizontally. But we need to know the total velocity. And the Pythagorean theorem comes to the rescue. Just square the 2 velocities, add them together, and take the square root. We already know the square is 490 from the work above, so sqrt(490+490) = sqrt(980) = 31.30495 m/s</span>
3 0
3 years ago
If an object is moving eastward and slowing down, then the direction of its acceleration vector is
iragen [17]
It would still be eastward because The direction of the velocity vector is always in the same direction as the direction which the object moves.

8 0
3 years ago
The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5
ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E =  K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) =  7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) =  4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.89e5)^{2}  + (7.89e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) net magnitude =  1.115e6\frac{N}{C} @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

 E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.88e5)^{2}  + (7.88e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) and (b) the net magnitude =  1.242e6\frac{N}{C} @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0
2 years ago
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