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KIM [24]
3 years ago
8

What is the sound intensity level in a car when the sound intensity is 0.525 μW/m2 ? Use I0 = 1.00×10−12 W/m2 for the reference

intensity.
Physics
1 answer:
never [62]3 years ago
8 0

Answer:

The sound intensity level in the car is 57.2 dB.

Explanation:

Sound intensity level in decibels, β = 10 log (I/I₀); where I = 0.525 × 10⁻⁶ W/m², I₀ = 1.0 × 10⁻¹² W/m²

β (dB) = 10 log ((0.525 × 10⁻⁶)/(1.0 × 10⁻¹²)) = 10 × 5.72 = 57.2 dB

Hope this Helps!!!

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If the 50-kg crate starts from rest and achieves a velocity of v = 4 m/s
Kay [80]

Answer:

P = 227 N

Explanation:

Assuming the crate is on horizontal ground and subject to a horizontal force.

F = ma

P - μmg = ma

P = m(a + μg)

P = m(v²/2s + μg)

P = 50(4²/(2(5))+ 0.3(9.8))

P = 227 N

6 0
3 years ago
What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?
kvv77 [185]

Answer:

W = 78.48N\\W =0.0784kN\\W = 8kgf\\

W= 3.6lbf\\W= 115.2 lbm*ft/s2

Explanation:

if

m=8kg=3.6lb\\

and g=9.81 m/s2=32.16 ft/s2

and

W=m*g

we can just replace de mass and gravity and we have

W = 8kg * 9.81 \frac{m}{s^{2} } =78.48N\\W = \frac{78.48N}{1000} =0.0784kN\\W = 8kgf\\

W= 3.6lbf\\W= 3.6lbm *32.16 ft/s2 =115.2 lbm*ft/s2

5 0
3 years ago
Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

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above the pavement.  That's where the head of the unsuspecting
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7 0
3 years ago
Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys
max2010maxim [7]

Answer:

The spring was compressed the following amount:

\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }

Explanation:

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E_i=E_f\\

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Kinetic Energy is calculated by KE= 1/2mv2
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