A crying baby emits sound with an intensity of 8.0 × 10-8 w/m2. calculate a reasonable estimate for the intensity level from a s
2 answers:
The intensity level from a set of quintuplets (five babies) : <u>56 dB</u>
<h3>Further explanation</h3>Wave intensity is the power of a wave that is moved through a plane of one unit that is perpendicular to the direction of the wave
Can be formulated
I = intensity, W m⁻²
P = power, watt
A = area, m²
The farther the distance from the sound source, the smaller the intensity
So the intensity is inversely proportional to the square of the distance from the source
Intensity level (LI) can be formulated
Io = 10⁻¹²
For the level of intensity of several sound sources as many as n pieces can be formulated:
<h3>LIn = LI1 + 10 log n </h3>
The intensity level of 1 baby is
LI = 10 log 8.10⁴
LI = 49
The intensity level of 5 babies :
LI5 = LI + 10 log n
LI5 = 49 + 10 log 5
LI5 = 49 + 7
LI5 = 56
<h3>Learn more</h3>The intensity of a laser beam
electric field
magnetism
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Answer:
4.0 m/s
Explanation:
The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.
Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is
where here we have
d = 3.0 m is the horizontal distance covered
vx is the horizontal velocity
t = 1.3 s is the duration of the fall
Solving for vx,
Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by
where
h = 4.0 m is the initial height
vy is the initial vertical velocity
We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy
So now we can find the magnitude of the initial velocity:
Answer:
Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]
Explanation:
We can solve this problem by using Newton's universal gravitation law.
In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m
Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.
Mathematically this equals:
When we match these equations the masses cancel out as the universal gravitational constant
To solve this equation we have to replace the first equation of related with the distances.
Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.
We work with positive value
rm = 38280860.6[m] = 38280.86[km]
<u>Second part</u>
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The distance between the Earth and this point is calculated as follows:
re = 3.84 108 - 38280860.6 = 345719139.4[m]
Now the acceleration can be found as follows:
Answer:
57,42 KJ
Explanation:
By a isobaric proces, the expresion for the works in the jpg adjunt. Then:
W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)
By the ideal gases law: PV=RTn
Then, in (1): (remember Pa = Pb)
W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)
Since we have 1 Kg air: How much is this in moles?
From bibliography: 28.96 g/mol
Then, in 1 Kg (1000 g) there are:
n = 34,53 mol
Finally, in (2):
W = (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ
