Answer:
28 cm and 32 cm
Explanation:
1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of
a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.
As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.
So, the minimum length is 28 cm and the maximum length is 32 cm.
Answer:
2
Explanation:
We know that in the Fraunhofer single-slit pattern,
maxima is given by

Given values
θ=2.12°
slit width a= 0.110 mm.
wavelength λ= 582 nm
Now plugging values to calculate N we get

Solving the above equation we get
we N= 2.313≅ 2
Answer:
a) I = 3.63 W / m²
, b) I = 0.750 W / m²
Explanation:
The intensity of a sound wave is given by the relation
I = P / A = ½ ρ v (2π f
)²
I = (½ ρ v 4π² s_{max}²) f²
a) with the initial condition let's call the intensity Io
cte = (½ ρ v 4π² s_{max}²)
I₀ = cte s² f₀²
I₀ = cte 10 6
If frequency is increase f = 2.20 10³ Hz
I = constant (2.20 10³) 2
I = cte 4.84 10⁶
let's find the relationship of the two quantities
I / Io = 4.84
I = 4.84 Io
I = 4.84 0.750
I = 3.63 W / m²
b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or
I = cte (f s)²
I = constant (0.250 10³ 4)²
I = cte 1 10⁶
the relationship
I / Io = 1
I = Io
I = 0.750 W / m²
Explanation:
using the formula: S=ut+½gt², where u=0, S=?, g=8m/s², t=10seconds.
S=ut+½gt² ("ut" term will cancel because u=0).
=> S= ½gt²
=>S = ½×8×10²
=>S = 4×100
=>S = 400m .
Therefore, the distance traveled by the body in 10s is 400m.
hope this helps you.