Ask question

Ask question

All categories

2 years ago

15

Find the intensity in decibels [i(db)] for each value of i. normal conversation: i = 106i0 i(db) = power saw a 3 feet: i = 1011i

0 i(db) = jet engine at 100 feet: i = 1018i0 i(db) =

2 answers:

White raven [17]2 years ago

8 0

Answer:

Normal Conversation: i=106i0

i(dB)=60

Power saw a 3 feet: i=1011i0

i(dB)=110

Jet engine at 100 feet: i=1018i0

i(dB)=180

Explanation:

if these are the same as edge, then these are the answers! :)

zalisa [80]2 years ago

6 0

Answer:

1. 60

2. 110

3. 180

You might be interested in

A puck of mass m = 0.085 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless str

Rina8888 [55]

Answer:

T = 11.93 N

Explanation:

Newton's second law to the puck in the circular path

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in radial direccion (N)

m : puck mass (kg)

a : radial acceleration of the puck (m/s²)

Data:

m = 0.085 kg

L = 0.72 m = R : radium of the circular path (m)

θ= one revolution = 2Π rad

t= 0.45 s

Angular speed of the puck

ω = θ/t

ω = 1 rev/0.45 s = (2π/0.45) rad/s

ω = 13.96 rad /s

Radial acceleration or centripetal

a = ω²*R

a = (13.96) ²* (0.72)

a = 140.3 m/s²

Magnitude of the tension in the string (T)

We apply the Formula (1)

∑F = m*a

T = (0.085 kg )* (140.3 m/s² )

T = 11.93 N

4 0

2 years ago

The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of

Novosadov [1.4K]

Answer:

The angle between the blue beam and the red beam in the acrylic block is

$\theta _d =0.19 ^o$

Explanation:

From the question we are told that

The refractive index of the transparent acrylic plastic for blue light is $n_F = 1.497$

The wavelength of the blue light is $F = 486.1 nm = 486.1 *10^{-9} \ m$

The refractive index of the transparent acrylic plastic for red light is $n_C = 1.488$

The wavelength of the red light is $C = 656.3 nm = 656.3 *10^{-9} \ m$

The incidence angle is $i = 45^o$

Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as

$r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

So

$r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ]$

$r_F = 28.18^o$

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

$r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

So

$r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ]$

$r_F = 28.37^o$

The angle between the blue beam and the red beam in the acrylic block

$\theta _d = r_C - r_F$

substituting values

$\theta _d = 28.37 - 28.18$

$\theta _d =0.19 ^o$

4 0

2 years ago

If a force always acts perpendicular to an object's direction of motion, that force cannot change the object's kinetic energy.

laiz [17]

This is very good conceptual question and can clear your doubts regarding work-energy theorem.
Whenever force is perpendicular to the direction of the motion, work done by that force is zero.
According to work-energy theorem,
Work done by all the force = change in kinetic energy.

here, work done = 0.
Therefore,
0=change in kinetic energy
This means kinetic energy remains constant.
Hope this helps

5 0

2 years ago

A typical garden hose has an inner diameter of 5/8". Let's say you connect it to a faucet and the water comes out of the hose wi

castortr0y [4]

Since speed (v) is in ft/sec, let's convert our diameters from inches to feet:
1) 5/8in = 0.625in
0.625in × 1ft/12in = 0.0521ft
2) 0.25in × 1ft/12in = 0.021ft
Equation:
$v = 4q \div ( {d}^{2} \pi) \: where \: q = flow \\ v = velocity \: (speed) \: and \: \\ d = diameter \: of \: pipe \: or \: hose \\ and \: \pi = 3.142$
$we \: can \: only \: assume \:that \\ flow \: (q) \:stays \: same \: since \: it \\ isnt \: impeded \: by \: anything \\ thus \:it \: (q)\: stays \: the \: same \: \\ so \: 4q \: can \: be \: removed \: from \: \\ the \: equation$
$then \: we \: can \: assume \: that \: only \\ v \: and \: d \: change \: leading \:us \: to > > \\ (v1 \times {d1}^{2} \pi) = (v2 \times {d2}^{2}\pi)$
$both \: \pi \: will \: cancel \: each \: other \: out \: \\ as \: constants \:since \: one \: is \: on \\ each \: side \: of \: the \: =$

$(v1 \times {d1}^{2}) = (v2 \ \times {d2}^{2}) \\ (7.0 \times {0.052}^{2}) = (v2 \times {0.021}^{2}) \\ divide \: both \: sides \: by \: {0.021}^{2} \\ to \: solve \: for \: v2 > >$
$v2 = (7.0)( {0.052}^{2} ) \div ( {0.021}^{2}) \\ v2 = (7.0)(.0027) \div (.00043) \\ v2 = 44 \: feet \: per \: second$
new velocity coming out of the hose then is
44 ft/sec

4 0

2 years ago

A spring has a spring constant of 48 N/m. The end of the spring hangs 8 m above the ground. How much weight can be placed on the

Setler79 [48]

The answer is 96 N .....................................

7 0

2 years ago

Read 2 more answers