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sineoko [7]
3 years ago
5

1. What is the intensity of a sound when an observer is 10.0 m from the speaker

Physics
1 answer:
Nana76 [90]3 years ago
8 0

Answer:

About 133 db.

Explanation:

Sound Intensity Level in db (SIL db) is equal to 10log (base 10) times the ratio of the sound intensity at 200 watts (I) relative to the sound intensity of  the reference sound intensity (I sub 0), which by default is equal to 10⁻¹² W/m² or 0 dB.

I = 200 w / 10 m^2 = 20 w per square meter  

I sub 0 = 10^-12 w per square meter

SIL = 10log ( I / I sub o) = 20 / 10^-12 = 10log ( 20^12) = 10 ( 13.3 ) = 133 db

Hope I typed this part correctly. Hard to get it in without being able to do exponents, etc. :D

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Compare lunar and solar eclipses and answer the following questions:
xeze [42]
So lunar eclips earth between sun and moon
Solar eclips moon between sun and earth.
About the 3th.. im not sure, it depends on if you meen a total solar eclips or not... i think total is more rare then a lunar eclipse..

3 0
3 years ago
A car accelerates from 30 m/s to 50 m/s in 2 seconds. Calculate the cars acceleration
masha68 [24]

Answer:

applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²

5 0
3 years ago
Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
DiKsa [7]

Answer:0.061

Explanation:

Given

T_C=300 k

Temperature of soup T_H=340 K

heat capacity of soup c_v=33 J/K

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any  instant

efficiency is given by

\eta =\frac{dW}{Q}=1-\frac{T_C}{T}

dW=Q(1-\frac{T_C}{T})

dW=c_v(1-\frac{T_C}{T})dT

integrating From T_H to T_C

\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT

W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT

W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}

W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]

Now heat lost by soup is given by

Q=c_v(T_C-T_H)

Fraction of the total heat that is lost by the soup can be turned is given by

=\frac{W}{Q}

=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}

=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}

=\frac{300-340-300\ln (\frac{300}{340})}{300-340}

=\frac{-40+37.548}{-40}

=0.061

4 0
3 years ago
Which part of the water cycle would follow step D in the diagram shown?
Dmitry [639]

Condensation.


iGreen, more like iNOTGreen amirite?

5 0
3 years ago
Read 2 more answers
A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parall
dolphi86 [110]

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

W_P=Px

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

W_P=(171N)(8.80m)=1504.8J

(b) The work don by the friction force is:

W_f=F_fx=\mu N x=\mu Mg x

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J

(c) The Normal force is

N=Mg=(46.0kg)(9,8m/s^2)=450.8N

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

W_N=0J

(d) the same as before:

W_g=0J

8 0
3 years ago
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