Answer:
string's damping is 1.03676
Explanation:
given data
sound level = 9.0 dB
time = 1 sec
to find out
string's damping
solution
we will apply here formula for string damping (b) that is
A(t) = A × 
...................1
we know here I ∝ A² so
√I(t) = √I ×
√I(t) / √I = .....................2
we know sound level decreases 9 dB i.e ΔdB = 9
so we can write
ΔdB = 10 log ( I(t) / I)
9 = 10 log ( I(t) / I)
I(t) / I = 
I(t) / I = 0.1258
and
√I(t) / I) = √0.1258 = 0.3546 .......................3
from equation 2 and 3 we get
0.3546 =
take ln both side
-bt = ln 0.3546
here we know t is 1 sec
so
- b = - 1.03676
b = 1.03676
so here string's damping is 1.03676
Answer:
fr = ½ m v₀²/x
Explanation:
This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.
The best way to solve this exercise is to use the energy work theorem
W = ΔK
Where work is defined as the product of force by distance
W = fr x cos 180
The angle is because the friction force opposes the movement
Δk =
–K₀
ΔK = 0 - ½ m v₀²
We substitute
- fr x = - ½ m v₀²
fr = ½ m v₀²/x
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
Answer:
A=0.199
Explanation:
We are given that
Mass of spring=m=450 g=
Where 1 kg=1000 g
Frequency of oscillation=
Total energy of the oscillation=0.51 J
We have to find the amplitude of oscillations.
Energy of oscillator=
Where 
=Angular frequency
A=Amplitude
Using the formula
Hence, the amplitude of oscillation=A=0.199
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