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3 years ago

7

Which example possesses mechanical potential energy?. A. a taut guitar string. B. an oscillating pendulum. C. a roller coaster r

ide. D. a child on a swing. E. a speeding car

2 answers:

Andrej [43]3 years ago

6 0

<span>
The taut guitar string haspotencial energy which we can see in action.</span> <span>· so option a is correct.</span>

777dan777 [17]3 years ago

5 0

a taut guitar string

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A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

<u>Part A</u>

I = 1.4 mW/m²

<u>Part B</u>

β = 91.46 dB

Explanation:

<u>Part A</u>

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

<u>Part B</u>

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

3 years ago

Steam enters a one-inlet, two-exit control volume at location (1) at 360°C, 100 bar, with a mass flow rate of 2 kg/s. The inlet

yaroslaw [1]

Answer:

The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

Explanation:

Given that,

Mass flow rate = 2 kg/s

Diameter of inlet pipe = 5.2 cm

Fifteen percent of the flow leaves through location (2) and the remainder leaves at (3)

The mass flow rate is

$m_{2}=0.15\times2$

We need to calculate the mass flow rate at reach exit

Using formula of mass

$m_{3}=m_{1}-m_{2}$

$m_{3}=2-0.15\times2$

$m_{3}=1.7\ kg/s$

We need to calculate the inlet velocity

Using formula of velocity

$v=\dfrac{m}{\rho A}$

Put the value into the formula

$v=\dfrac{2}{42.868\times\dfrac{\pi}{4}\times(5.2\times10^{-2})^2}$

$v=21.9\ m/s$

Hence, The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

7 0

3 years ago

33] You have long hair, you should:

Sveta_85 [38]

Answer:

B

Explanation:

hair can be a safety hazard

4 0

3 years ago

Read 2 more answers

20% Part (a) Use an "E Field Sensor" and move it along either equipotential. What can you say about the E field along an equipot

Alex

Answer:

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

Explanation:

As we know that the relation between electric field and electric potential is given as

$\Delta E = -\frac{dV}{dr}$

here if we say that potential is constant because electric field sensor is moving along equi-potential line.

Then we will say

V = constant

so we have

$\Delta E = 0$

so electric field will remain constant always in magnitude and always remains perpendicular to the surface

so we have

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

6 0

3 years ago

A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck trav

aniked [119]

<span>You can use the equation
V_xf = V_xi + a_x(t)

V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0 t

Thus, solve for t and get 10seconds and then take 5 seconds to break after 20 seconds of driving so for

a) 10 + 20 + 5 = 35 seconds

</span><span>for part b)
You can use the formula
Delta x/Delta t = average velocity

Need to find xf, knowing xi = 0

Thus, use the formula
x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
x_f = 100m

so for the first 10 seconds the truck traveled 100ms At a speed of 20m/s

20m/s = xm/20s 20*20 = x
x = 400

thus we have 100+400 = 500m then it slows down from 500m to x_f

thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
x_f = 500 + (1/2)(0 + 20)(5) x_f = 500 + 50
x_f = 550

therefore the total distance traveled is 550m
</span>
<span>to calculate average velocity
550/35 = 16m/s

thus V_xavg = 16m/s</span>

8 0

3 years ago