Disregarding air friction, what force(s) act on a bullet shot from a rifle as it moves through the air?
1 answer:
The gravity.
In fact, the gravity will "pull" the bullet towards the ground, and the magnitude of the force is equal to the weight of the bullet:
where
m is the mass of the bullet
is the acceleration of gravity
Due to the presence of this force, the bullet will have a parabolic motion, which consists of two independent motions on the horizontal axis and on the vertical axis:
- on the horizontal axis, the bullet moves by uniform motion with constant speed
- on the vertical axis, the bullet moves by uniformly accelerated motion, with constant acceleration g towards the ground.
In fact, the gravity will "pull" the bullet towards the ground, and the magnitude of the force is equal to the weight of the bullet:
where
m is the mass of the bullet
Due to the presence of this force, the bullet will have a parabolic motion, which consists of two independent motions on the horizontal axis and on the vertical axis:
- on the horizontal axis, the bullet moves by uniform motion with constant speed
- on the vertical axis, the bullet moves by uniformly accelerated motion, with constant acceleration g towards the ground.
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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