Answer:
r₂ = 0.316 m
Explanation:
The sound level is expressed in decibels, therefore let's find the intensity for the new location
β = 10 log 
let's write this expression for our case
β₁ = 10 log \frac{I_1}{I_o}
β₂ = 10 log \frac{I_2}{I_o}
β₂ -β₁ = 10 ( 
)
β₂ - β₁ = 10 
log \frac{I_2}{I_1} = 
= 3
= 10³
I₂ = 10³ I₁
having the relationship between the intensities, we can use the definition of intensity which is the power per unit area
I = P / A
P = I A
the area is of a sphere
A = 4π r²
the power of the sound does not change, so we can write it for the two points
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
we substitute the ratio of intensities
I₁ r₁² = (10³ I₁ ) r₂²
r₁² = 10³ r₂²
r₂ = r₁ / √10³
we calculate
r₂ = 
r₂ = 0.316 m
Answer: The distance is 723.4km
Explanation:
The velocity of the transverse waves is 8.9km/s
The velocity of the longitudinal wave is 5.1 km/s
The transverse one reaches 68 seconds before the longitudinal.
if the distance is X, we know that:
X/(9.8km/s) = T1
X/(5.1km/s) = T2
T2 = T1 + 68s
Where T1 and T2 are the time that each wave needs to reach the sesmograph.
We replace the third equation into the second and get:
X/(9.8km/s) = T1
X/(5.1km/s) = T1 + 68s
Now, we can replace T1 from the first equation into the second one:
X/(5.1km/s) = X/(9.8km/s) + 68s
Now we can solve it for X and find the distance.
X/(5.1km/s) - X/(9.8km/s) = 68s
X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s
X = 68s/0.094s/km = 723.4 km
answers 93 bar
The atmosphere of Venus is the layer of gases surrounding Venus. It is composed primarily of carbon dioxide and is much denser and hotter than that of Earth. The temperature at the surface is 740 K (467 °C, 872 °F), and the pressure is 93 bar (9.3 MPa), roughly the pressure found 900 m (3,000 ft) underwater on Earth.
The answer is C. Hope this helps.
Set up the problem with the conversion rates as fractions where when you multiply the units cancel out leaving the desired units behind.
