The nonreflective coating on a camera lens with an index of refraction of 1.21 is designed to minimize the reflection of 570-nm
1 answer:
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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 value is
Explanation:
From the question we are told that
The velocity which the rover is suppose to land with is
The mass of the rover and the parachute is
The drag coefficient is
The atmospheric density of Earth is
The acceleration due to gravity in Mars is
Generally the Mars atmosphere density is mathematically represented as
=>
=>
Generally the drag force on the rover and the parachute is mathematically represented as
=>
=>
Gnerally this drag force is mathematically represented as
Here A is the frontal area
So
=>
=>