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The wavelength of the note is 
. Since the speed of the wave is the speed of sound, 
, the frequency of the note is
Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
where
is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
where
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
Answer:
a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º
This is 8º north west
Explanation:
This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately
To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.
First vector A = 2 to 20º north west
Measured from the positive x axis is θ = 180 -20 = 160º
We use trigonometry to find the components
Cos 20 = Aₓ / A
sin 20 = / A
Aₓ = A cos 160 = 2 cos 160
= A sin160 = 2 sin160
Aₓ = -1,879 mi
= 0.684 mi
Second vector B = 4 mi 10º west of the south
Angle θ = 270 - 10 = 260º
cos 2600 = Bₓ / B
sin 260 = / B
Bₓ = B cos 260
= B sin 260
Bₓ = 4 cos 260
= 4 sin 260
Bₓ = -0.6946mi
= - 3,939 mi
Third vector C = 3 mi to 15 north east
cos 15 = Cₓ / C
sin15 = / C
Cₓ = C cos 15
= C sin15
Cₓ = 3 cos 15
= 3 sin 15
Cₓ = 2,898 mi
= 0.7765 mi
Now we can find the final position of the person
X = Aₓ + Bₓ + Cₓ
X = -1.879 -0.6949 + 2.898
X = 0.3241 mi
Y = +
+
Y = 0.684 - 3.939 +0.7765
Y = -2.4785 mi
a) We use Pythagoras' theorem
R = √ (x2 + y2)
R = √ (0.3241 2 + (-2.4785) 2)
R = 2.4996 mi
R = 2.5 mi
b) let's use trigonometry
Tan θ = y / x
Tanθ = -2.4785 / 0.3241
θ = tan⁻¹ (-7,647)
θ = -82
Measured from the positive side of the x axis is Te = 360 - 82 = 278º
(90-82) south east
To return to your case you must walk in the opposite direction or Te = 98º
This is 8º north west
Answer:
8.91 J
Explanation:
mass, m = 8.20 kg
radius, r = 0.22 m
Moment of inertia of the shell, I = 2/3 mr^2
= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2
n = 6 revolutions
Angular displacement, θ = 6 x 2 x π = 37.68 rad
angular acceleration, α = 0.890 rad/s^2
initial angular velocity, ωo = 0 rad/s
Let the final angular velocity is ω.
Use third equation of motion
ω² = ωo² + 2αθ
ω² = 0 + 2 x 0.890 x 37.68
ω = 8.2 rad/s
Kinetic energy,
K = 0.5 x 0.265 x 8.2 x 8.2
K = 8.91 J