Rocket thrust equation
= ( mass flow rate of fuel burnt ) X (Velocity of gas ejected ) + ( Exit Pressure - Outdoor Pressure ) X ( Area of exhaust )
In this case, we can assume the exit pressure = outdoor pressure and since area of exhaust is not given, it can be assumed to be negligible.
In this case, by Newton 3rd’s law,
Force exerted by gas on rocket
= Force exerted by rocket on gas
= (10kg/s) X (5 x 10^3 m/s)
= 5 x 10^4 N
Answer:
Incident ray
Explanation:
"To describe the reflection of light, we will use the following terminology. The incoming light ray is called the incident ray. The light ray moving away from the surface is the reflected ray. The most important characteristic of these rays is their angles in relation to the reflecting surface."
https://www.siyavula.com/read/science/grade-11/geometrical-optics/05-geometrical-optics-03
Answer:
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Answer:
are sure this is a question
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
