What type of energy does an object gain as it is lifted at a constant speed?

A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be for a

andriy [413]

To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

$sin\theta = 1.22\frac{\lambda}{d}$

Here,

$\lambda$ = Wavelength

d= Diameter of aperture

$\theta$ = Angular resolution or diffraction angle

Our values are given as,

$\theta = 11\°$

The frequency of the sound is $f = 9100 Hz$

The speed of the sound is $v = 343 m/s$

The wavelength of the sound is

$\lambda = \frac{v}{f}$

Here,

v = Velocity of the wave

f = Frequency

Replacing,

$\lambda = \frac{(343 m/s)}{(9100 Hz)}$

$\lambda = 0.0377 m$

The diffraction condition is then,

$sin\theta = 1.22\frac{\lambda}{d}$

Replacing,

$sin(11\°) = 1.22\frac{(0.0377 m)}{(d)}$

d = 0.24 m

Therefore the diameter should be 0.24m

6 0

3 years ago

You stand17.5 m from a wall holding a baseball. You throw the baseball at the wall at an angle of 20.5∘ from the ground with an

Lelechka [254]

Answer: 8.8 m

Explanation:

The movement of the baseball to the wall is a example of parabolic motion. While the baseball aproach the wall it is affected by gravity.

In this case, because the Initial Velocity of the ball is a vecotr, it can be defined using its two directionals compounds. One, on the Y-axis and another on the X-axis. This can be related, on how a hypotenuse is the product of two legs of the triangle. Because of this, each one of this, can be know using the following equation:

Vx = Vo * cos(∅)

Vy = Vo * sin(∅)

Where Vo is the initial velocity 27.5 m/s, and ∅ is the angle which is 20.5°. So we calculate:

Vx = 27.5m/s * cos(20.5)

Vx = 27.5m/s * 0.936

Vx = 25.74m/s

Vy = 27.5m/s * sin(20.5)

Vy = 9.62m/s

Now the movement is divided on two parts, one under the effect of gravity and another one with a constant velocity.

To know how tall does the baseball hit the wall, we need to know first how much time it takes the ball to reach the wall on the X-axis. The wall is 17.5m away, we velocity on Vx that is constant we can calculate as it follow:

Time (T) = Distance (D) / Velocity (V)

Where Distance is 17.5m and our Velocity is the Vx calculated before.

T = 17.5 m / 25.74m/s

T = 0.68s

This is the time it takes the ball to reach the wall.

Know with the time, we can calculate the how tall it got on that time with the following equation:

$x = (Vo*t) + (\frac{1}{2} *a*t^{2} )$

Where Vo is the Y-Compound of the Initial Velocity.

a is the aceleration, in this case the Gravity. Which, will be negative because is oposing the movement. Gravity is equal to 9.81 $m/s^{2}$

And t is the time it takes the ball to get to the wall.

$x = (Vy*T) + (\frac{1}{2} *g*T^{2} )$

$x = ((9.62m/s)*0.68s) + (\frac{1}{2} *(9.81m/s^{2})*(0.68s)^{2} )$

$x = 8.8 m$

This is the height that the baseball touch the wall.

6 0

3 years ago

Help me please, it's so hard

fomenos

Answer:

4.58×10²³ atoms

5.94×10⁻²¹ J

1340 m/s

Explanation:

Use ideal gas law to find moles of gas.

PV = nRT

(1.266 atm × 101300 Pa/atm) (4/3 π (0.15 m)³) = n (8.31451 J/mol/K) (14 + 273) K

n = 0.760 mol

Use Avogadro's number to find number of atoms.

(0.760 mol) (6.02214×10²³ atom/mol) = 4.58×10²³ atoms

Average kinetic energy per molecule is:

KE = 3/2 kT

KE = 3/2 (1.38066×10⁻²³ J/K) (14 + 273) K

KE = 5.94×10⁻²¹ J

RMS speed of each atom is:

KE = 1/2 mv²

5.94×10⁻²¹ J/atom = 1/2 (0.004 kg/mol) (1 mol / 6.02214×10²³ atom) v²

v = 1340 m/s

8 0

3 years ago

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Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the t

coldgirl [10]

The question is incomplete. The complete question is :

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension? The applied tensile force is 102 KN. The unloaded cube edge length is 56 mm. The glass fibers have an elastic modulus of 200 GPa. The epoxy has an elastic modulus of 38 GPa. The cube is comprised of 18 vol% epoxy (the balancing vol % is glass fiber). Hint: The loading axis is intentionally unspecified. Answer Format: Lowest possible length increase (change of length) under tension.

Solution :

Given :

$E_{glass fibre}$ = 200 GPa

$V_{glass fibre} = 82\%$

$E_{epoxy}$ = 38 GPa

$V_{epoxy} = 82\%$

Edge length = 56 mm

Cube is loaded in axial tension such that the force is uniformly applied over a cube face.

$E_{\text{composite}}=\frac{E_{glass fibre} \times E_{epoxy}}{(E_{glass fibre .E_{epoxy}})+(E_{fibre}.V_{glass fibre})}$

$E_{composite} = \frac{200 \times 38}{(200 \times 0.18)+(38\times 0.82)}$

$=113.16$ GPa

Applied stress $=\frac{\text{applied load}}{\text{area}}$

$\sigma=\frac{102 \times 10^3 \ N}{56 \times 56 \times 10^{-6} \ m^2}$

= 32.5 MPa

By Hooke's law

$\sigma = E . \epsilon$

$\sigma = E. \frac{\Delta l}{l}$

$\Delta l = \frac{\sigma}{E}\times l$

Length change, $\Delta l =\frac{32.5 \times 10^6 \ Pa}{113.16 \times 10^9 \ Pa}\times 56 \times 10^{-2} \ m$

$\Delta l = \frac{32.5 \times 56}{113.16} \times 10^{-3} \ mm$

= 0.016 mm

7 0

3 years ago

A construction worker pushes a cement block with a 750 N force for 10 minutes without the cement block moving. How much work did

lutik1710 [3]

The work done by a system on a different body is equal to the product of the force exerted and the distance that the body has move in parallel to the force exerted. In this item, we have to determine first the distance and multiply it with the given force equal to 750N.

8 0

3 years ago

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