Given: <span>ω(t) = 8.00 - 0.2*t^2
Differentiating both sides with respect to time we get:
angular acceleration = </span>α(t) = -0.2 * 2 *t
At t = 6.8 sec,
angular acceleration = α(t) = -0.2 * 2 *6.8 = -2.72 rad/s^2
<u>Answer</u>
a. tension (associated with normal faults)
c. diagonal
<u>Explanation</u>
A pull of spring or of string on both ends of an object is called tension. So for the question one, the answer is <em> tension (associated with normal faults)</em>
There are three types of faults. Faults are produced by stress or strain by moving plates. These faults are: normal faults, reverse faults and transcurrent or Strike-slip. Strike-slip faults can also be called transform fault. The answer to the second question is <em>c. diagonal.</em>
Answer:
93.54 Hz
Explanation:
✓From the question, Number of harmonic frequency is 4
✓ the frequency (f₄ )= 116.5 Hz
✓harmonic frequency can be calculated using below expresion
fₙ = [ (nv)/4L]..........eqn(1)
v = speed of sound= 343 m/s
n = number of given harmonic frequency
L = Length of the rope
Using above expresion ,and substitute the values at (n=4) which is 4th harmonic frequency to find the " initial Lenght of the rope
fₙ = [ (nv)/4L]
f₄ = 4× 343 /4L
f₄ = 343 /L
L= 343 /f₄
But f₄= 116.5 Hz
L= 343/116.5= 2.944m
Hence, initial Lenght of the rope= 2.944m
We can determine the frequency of new length as ( initial Lenght of the rope + tubing Lenght)
= ( 2.944m + 0.721m )
= 3.667m
Hence, new length= 3.667m
To find the new frequency of the 4th harmonic we will use eqn(2)
f₄ = v/l ...............eqn(2)
From equation (2) If we substitute the values we have
f₄ = (343/3.667)
= 93.54 Hz
Hence, the the new frequency of the 4th harmonic is
93.54 Hz
Answer:
6400 m
Explanation:
You need to use the bulk modulus, K:
K = ρ dP/dρ
where ρ is density and P is pressure
Since ρ is changing by very little, we can say:
K ≈ ρ ΔP/Δρ
Therefore, solving for ΔP:
ΔP = K Δρ / ρ
We can calculate K from Young's modulus (E) and Poisson's ratio (ν):
K = E / (3 (1 - 2ν))
Substituting:
ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)
Before compression:
ρ = m / V
After compression:
ρ+Δρ = m / (V - 0.001 V)
ρ+Δρ = m / (0.999 V)
ρ+Δρ = ρ / 0.999
1 + (Δρ/ρ) = 1 / 0.999
Δρ/ρ = (1 / 0.999) - 1
Δρ/ρ = 0.001 / 0.999
Given:
E = 69 GPa = 69×10⁹ Pa
ν = 0.32
ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)
ΔP = 64.0×10⁶ Pa
If we assume seawater density is constant at 1027 kg/m³, then:
ρgh = P
(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa
h = 6350 m
Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.
Answer:
temperature of about 2.72548 ± 0.00057 K.
Explanation:
The cosmic background radiation is an electromagnetic radiation that remains from an early stage of the universe during the big bang. It was accidentally discovered in 1965 by two American radio astronomers Arno Penzias and Robert Wilson. The radiation was given off before the formation of stars and planets, when the universe was young, denser, hotter, and filled with a uniform glow from a white-hot fog of hydrogen plasma, which cooled down as the universe expanded. From calculations, it was deduced that the radiation had a temperature of 2.72548±0.00057 K, which is close to the temperature of the universe during its formation.
