A man walks 30 m to the west, then 5 m to the east in 45 seconds.

What is the speaker’s power output if the sound intensity level is 102 dBdB at a distance of 25 mm ? Express your answer to two

lesya [120]

Answer:

Power = 124.50 W

Explanation:

Given that:

The Sound intensity of a speaker output is 102 dB

and the distance r = 25 m

For the intensity of sound,

$\beta (dB)= 10 \ log_{10 } (\dfrac{I}{I_o})$

where;

the threshold of hearing $I_o = 10^{-12} (W/m^2)$

$\dfrac{102 }{10}= log_{10}( \dfrac{I}{10^{-12}})$

$10^{10.2} = \dfrac{I}{10^{-12}}$

$I = 10^{10.2} \times 10^{-12}$

I = 0.01585 W/m²

If we recall, we know remember that ;

Power = Intensity × A rea

Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²

Power = 124.50 W

7 0

3 years ago

A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

emmasim [6.3K]

• Net vertical force on the block:

∑ F = n - w = 0

(n = magnitude of normal force, w = weight)

n = w = m g

(m = mass, g = 9.8 m/s²)

n = (4 kg) (9.8 m/s²) = 39.2 N

• Net horizontal force:

∑ F = -f = m a

(f = mag. of friction, a = acceleration)

We have f = µ n = 0.5 (39.2 N) = 19.6 N, so

-19.6 N = (4 kg) a

a = -4.9 m/s²

With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆x in this time such that

0² - (5 m/s)² = 2 (-4.9 m/s²) ∆x

∆x = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m

8 0

3 years ago

If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c

Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction $\mu =0.8$

Angle of inclination is equal to $\Theta =tan^{-1}\mu$

$\Theta =tan^{-1}0.8=38.65^{\circ}$

$tan{38.65^{\circ}}=0.8$

Radius is given r = 28 m

Acceleration due to gravity $g=9.8m/sec^2$

We know that $tan\Theta =\frac{v^2}{rg}$

$0.8=\frac{v^2}{28\times 9.8}$

$v^2=219.52$

$v=14.816m/sec$

So before start of slide velocity will be 14.81 m/sec

3 0

3 years ago

Menciona generaciones de particulas y explique cada una

Readme [11.4K]

ENGLISH:

It distinguishes between the fermions, which are particles of matter, and the bosons, which carry forces. The matter particles include six quarks and six leptons. The six quarks are called the up, down, charm, strange, top and bottom quark. ... All of these matter particles fall into three “generations.”

SPANISH:

Distingue entre los fermiones, que son partículas de materia, y los bosones, que transportan fuerzas. Las partículas de materia incluyen seis quarks y seis leptones. Los seis quarks se denominan quark arriba, abajo, encanto, extraño, superior e inferior. ... Todas estas partículas de materia se dividen en tres "generaciones".

3 0

3 years ago

Which of the following metals require ultraviolet light to exhibit the photoelectric effect?The options available: a. Cs, work f

Eddi Din [679]

Answer:

b. AG, work function=4.74eV

Explanation:

Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:

$\lambda=380 nm =3.8\cdot 10^{-7}m$ (wavelength of lowest-energy ultraviolet light)

So, the lowest energy of ultraviolet light can be found by using the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

Substituting,

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.8\cdot 10^{-7} m}=5.23\cdot 10^{-19}J$

And keeping in mind that

$1 eV = 1.6\cdot 10^{-19}J$

This energy converted into electronvolts is

$E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV$

The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:

b. AG, work function=4.74eV

Because for all the other metals, visible light will be enough to extract photoelectrons.

7 0

3 years ago