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enyata [817]
1 year ago
11

Q|C A sinusoidal wave in a string is described by the wave function

Physics
1 answer:
creativ13 [48]1 year ago
4 0

The maximum transverse force on a 1.00cm segment of the string is 0.045N.

The maximum transverse force on the segment,

l=1.00cm=1.00cm×1m/100cm=0.01m is,

F=μla

=(12.0×10^−3kg/m)×(0.01m)×(375m/s^2)

=0.045N

Hence the maximum force is 0.045N

Learn more about transverse force here:

brainly.com/question/13618693

#SPJ4

You might be interested in
The length of a simple pendulum is 0.66 m, the pendulum bob has a mass of 310 grams, and it is released at an angle of 12 degree
lina2011 [118]
A) the periodic time is given by the equation;
 T= 2π√(L/g)
For the frequency will be obtained by 1/T (Hz)
T = 2 × 3.14 √ (0.66/9.81)
   = 6.28 × √0.0673
    = 1.6289 Seconds
Frequency = 1/T = f = 1/1.6289
 thus; frequency = 0.614 Hz

b)  The vertical distance, the height is given by
 h= 0.66 cos 12
 h = 0.65 m
Vertical fall at the lowest point = 0.66 - 0.65 = 0.01 m
Applying conservation of energy
energy lost (MgΔh) = KE gained (1/2mv²)
 mgh = 1/2mv²
  v² = 2gΔh = 2×9.81 × 0.01 
                   = 0.1962
v = 0.443 m/s

c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
Thus total energy is equal to;
E = 1/2mv²
   = 1/2 × 0.310 × 0.443²
   = 0.0304 J


4 0
3 years ago
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. Wha
ikadub [295]

We have that the spring constant is mathematically given as

k=2.37*10^{11}N/m

Generally, the equation for angular velocity is mathematically given by

\omega=\sqrt{k}{m}

Where

k=spring constant

And

\omega =\frac{2\pi}{T}

Therefore

\frac{2\pi}{T}=\sqrt{k}{n}

Hence giving spring constant k

k=m((\frac{2 \pi}{T})^2

Generally

Mass of earth m=5.97*10^{24}

Period for on complete resolution of Earth around the Sun

T=365 days

T=365*24*3600

Therefore

k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2

k=2.37*10^{11}N/m

In conclusion

The effective spring constant of this simple harmonic motion is

k=2.37*10^{11}N/m

For more information on this visit

brainly.com/question/14159361

8 0
3 years ago
the first s-wave reaches a seismic station 22 minutes after an earthquake occurred. how long did it take the first p-wave to rea
Naddik [55]

The time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.

<h3>Time of travel of the P-wave</h3>

In rock, S waves generally travel about 60% the speed of P waves, and the S wave always arrives after the P wave.

<h3>Relationship between speed and time</h3>

v ∝ 1/t

v₁t₁ = v₂t₂

t₁/t₂ = v₂/v₁

t₁/t₂ = 0.6v₁/v₁

t₁/t₂ =  0.6

t₁ = 0.6t₂

t₁ = 0.6 x 22 mins

t₁ = 13.2 mins

Thus, the time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.

Learn more about P-waves here: brainly.com/question/2552909

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7 0
2 years ago
A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th
Mila [183]

Answer:

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0

\frac{3}{(r + a)^2} = \frac{2}{r^2}

\frac{r+a}{r} = \sqrt{\frac{3}{2}}

1 + \frac{a}{r} =\sqrt{\frac{3}{2}}

\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}

so we will have

r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}

so the x coordinate of this position is given as

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

6 0
3 years ago
What is the answer for the question -6 + (-8) =
Vaselesa [24]
The answer is negative 14
6 0
3 years ago
Read 2 more answers
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