I = pressure amplitude given = 0.2 W/m²
dB = decibel reading
decibel reading from the pressure amplitude is given as
dB = 10 log₁₀ (I/10⁻¹²)
inserting the values in the above equation
dB = 10 log₁₀ (0.2/10⁻¹²)
dB = 10 log₁₀ (2 x 10⁻¹/10⁻¹²)
dB = 10 log₁₀ (2 x 10⁻¹.10¹²)
dB = 10 log₁₀ (2 x 10¹²⁻¹)
dB = 10 log₁₀ (2 x 10¹¹)
dB = 113.01 db
hence the decibel reading comes out to be 113.01 db
Answer:
v_{f} = 74 m/s, F = 230 N
Explanation:
We can work on this exercise using the relationship between momentum and moment
I = ∫ F dt = Δp
bold indicates vectors
we can write this equations in its components
X axis
Fₓ t = m ( -v_{xo})
Y axis
t = m (v_{yf} - v_{yo})
in this case with the ball it travels horizontally v_{yo} = 0
Let's use trigonometry to write the final velocities and the force
sin 30 = v_{yf} / vf
cos 30 = v_{xf} / vf
v_{yf} = vf sin 30
v_{xf} = vf cos 30
sin40 = F_{y} / F
F_{y} = F sin 40
cos 40 = Fₓ / F
Fₓ = F cos 40
let's substitute
F cos 40 t = m ( cos 30 - vₓ₀)
F sin 40 t = m (v_{f} sin 30-0)
we have two equations and two unknowns, so the system can be solved
F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)
F sin 40 0.1 = 0.4 v_{f} sin 30
we clear fen the second equation and subtitles in the first
F = 4 sin30 /sin40 v_{f}
F = 3.111 v_{f}
(3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80
v_{f} (3,111 cos 40 -4 cos30) = - 80
v_{f} (- 1.0812) = - 80
v_{f} = 73.99
v_{f} = 74 m/s
now we can calculate the force
F = 3.111 73.99
F = 230 N
I believe it wattage or watts
Explanation:
For the equilibrium:
\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx=0ρ
wood
gh−ρ
oil
g(h−x)−ρ
water
gx=0
\rho_{wood}h-\rho_{oil}(h-x)-\rho_{water}x=0ρ
wood
h−ρ
oil
(h−x)−ρ
water
x=0
(974)(3.97)-928(3.97-x)-1000x=0(974)(3.97)−928(3.97−x)−1000x=0
x=2.54\ cmx=2.54 cm
Explanation:
Doing homework is risky behaviour broo
