A string that is 3.6 m long is tied between two posts and plucked. The string produces a wave that has a frequency of 320 Hz and
1 answer:
To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as
Where,
v = Velocity
f = Frequency,
Our values are given as
L = 3.6m
v= 192m/s
f= 320Hz
Replacing we have that
The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,
Therefore the number of wavelengths of the wave fit on the string is 6.
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Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final jeight
is the bomb'e initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's fina velocity
Knowing this, let's begin with the answers:
<h3>b) Time</h3>With the conditions given above, equation (1) is now written as:
(4)
Isolating :
(5)
(6)
(7)
<h3>a) Final velocity</h3>
Since , equation (3) is written as:
(8)
(9)
(10) The negative sign ony indicates the direction is downwards
<h3>c) Range</h3>
Substituting (7) in (2):
(11)
(12)
Answer:
1.1397 Nm
Explanation:
When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth.
If the muscle generates a force
and
, then the torque is equal to
we see that r = 2.65 cm = 0.0265 m
therefore
torque = 0.0265 x 49.5
= 1.1397 Nm
Answer:
The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
Given that,
dielectric constant k = 5.5
the area of each plate, A = 0.034 m²
separating distance, d = 2.0 mm = 2 x 10⁻³ m
magnitude of the electric field = 200 kN/C
Capacitance of the capacitor is calculated as follows;
Maximum potential difference:
V = E x d
V = 200000 x 2 x 10⁻³ = 400 V
Maximum energy that can be stored in the capacitor:
E = ¹/₂CV²
E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²
E = 6.62 x 10⁻⁵ J
Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J