A bowling ball of 35.2 kg, generates 218 kg* m/s units of momentum. What is the velocity of the ball?
1 answer:
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Answer:
0.453 m/s
Explanation:
Assuming the handle has diameter of 0.4 m while inner part diameter is 0.1 m then the circumference of outer part is where d is diameter and subscript h denote handle. By substituting 0.4 for the handle's diameter then cirxumference of outer part is
The rate of rotation will then be 1.81/1.256=1.441 rev/s
Similarly, circumference of inner part will be where subscript i represent inner. Substituting 0.1 for inner diameter then
The rate of rotation found for outer handle applies for inner hence speed will be 0.3142*1.441=0.453 m/s
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