Answer:
The amplitude of the eardrum's oscillation is 6.65×10^-13 m.
Explanation:
Given data:
The sound has a frequency of 262 Hz
The sound level is 84 dB
The air density is 1.21 kg/m^3
The speed of sound is 346 m/s
Solution:
As, Intensity of sound is given by,
I = Io×10^(s/10 db)
I = 2×π^2×ρ×v×f^2×Sm^2
Thus,
Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)
Now, put the values,
Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )
Sm = √(2.51×10^-4 / 5.66×10^8)
Sm = √0.443×10^-12
Sm = 6.65×10^-13 m.
That's a velocity.
It has a speed and the direction of the speed.
Answer:
u/2 √(1 + 3 cos² θ)
Explanation:
The object is thrown at an angle θ, so the velocity has two components, vertical and horizontal.
Initially, the vertical component is u sin θ and the horizontal component is u cos θ.
At the maximum height, the vertical component is 0 and the horizontal component is u cos θ.
The mean vertical velocity is:
(u sin θ + 0) / 2 = u/2 sin θ
The mean horizontal velocity is:
(u cos θ + u cos θ) / 2 = u cos θ
The net mean velocity can be found with Pythagorean theorem:
v² = (u/2 sin θ)² + (u cos θ)²
v² = u²/4 sin² θ + u² cos² θ
v² = u²/4 (1 − cos² θ) + u² cos² θ
v² = u²/4 (1 − cos² θ) + u²/4 (4 cos² θ)
v² = u²/4 (1 − cos² θ + 4 cos² θ)
v² = u²/4 (1 + 3 cos² θ)
v = u/2 √(1 + 3 cos² θ)
answer:projectile motion
Explanation:
Since gravity acts vertically, there is no acceleration in the horizontal (x) direction. This special type of two-dimensional motion is called projectile motion.
Explanation:
It depends because the higher the object is, the bigger the shadow, however, the object should be close to the line directly from the sun, for you to see the shadow visible.
