Let loudness be L, distance be d, and k be the constant of variation such that the equation that would best represent the given above is,
L = k/(d^2)
For Case 1,
L1 = k/(d1^2)
For Case 2,
L2 = k/((d1/4)^2)
For k to be equal, L1 = 16L2.
Therefore, the loudness at your friend's position is 16 times that of yours.
Answer:
= 829.69 Watt
≅ 830 Watt
Explanation:
Given that,
Velocity of air flow = 12.5m/s
Rate of flow of air = 9m³/s
Density of air = 1.18kg/m³
power by kinetic energy = 1/2(mv²)
mass = density × volume
m = 1.18 × 9
= 10.62 kg/s
power = 1/2 mV²
= 1/2 (10.62 × 12.5²)
= 829.69 Watt
≅ 830 Watt
Flow rate
u
=
9
m
3
/
s
Velocity of the air
V
=
8
m/s
Density of the air
ρ
=
1.18
kg
/
m
3
Answer:
Final speed of the crate is 15 m/s
Explanation:
As we know that constant force F = 80 N is applied on the object for t = 12 s
Now we can use definition of force to find the speed after t = 12 s
so here we know that object is at rest initially so we have
Now for next 6 s the force decreases to ZERO linearly
so we can write the force equation as
now again by same equation we have
put t = 6 s
Answer:
160N
Explanation:
Moments must be conserved - so.
C I believe is the correct answer. Developing possible solutions would be easier than spending hours researching or identifying the need.
