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Ludmilka
2 months ago
13

Find the three longest wavelengths (call them λ1, λ2, and λ3) that "fit" on the string, that is, those that satisfy the boundary

conditions at x=0 and x=l. these longest wavelengths have the lowest frequencies. express the three wavelengths in terms of l. list them in decreasing order of length, separated by commas.

Physics
1 answer:
Sav [3.1K]2 months ago
6 0
For a string that is secured at both ends, the oscillation amplitude must be zero at both extremities of the string. Refer to the attached images for clarity. It is evident that our fundamental harmonic will have the following wavelength: All subsequent harmonics are simply multiples of the fundamental. The three longest wavelengths are as follows:
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An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.51 μC/m2. Another infini
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E_total = 5.8 x 10⁴ N/C Explanation: To determine the electric field at specified points, we must calculate the vectors individually for each charge and sum them. The electric field caused by each charged conductive sheet can be derived via Gauss's law with the understanding of scalar products between the electric field and relevant surfaces.
4 0
2 months ago
A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
ValentinkaMS [3465]

Response:

83%

Clarification:

At the surface, the weight can be expressed as:

W = GMm / R²

where G denotes the gravitational constant, M represents the Earth's mass, m signifies the shuttle's mass, and R is the Earth's radius.

When in orbit, the weight is given by:

w = GMm / (R+h)²

where h indicates the shuttle's altitude above Earth's surface.

The weight ratio is as follows:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

For R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

Thus, the shuttle maintains 83% of its weight as it orbits.

4 0
2 months ago
A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri
Maru [3345]
The charge on the plastic cube is determined as follows.
7 0
2 months ago
THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A
ValentinkaMS [3465]

Answer: total angular distance = 1700° and 29.7 rad

   the total angular displacement = 0

Explanation:

This is a breakdown of the calculations needed.

The task is to determine both the total angular distance and displacement experienced by the knee.

To calculate the distance traveled by the knee, consider that when squatting, the knee bends 85° to lower, and then another 85° to return to standing (upright). Thus, the cumulative angular movement during the squat totals 170°.

For 10 squats, the knee must undergo 170° motion multiplied by 10, resulting in:

10 * 170° = 1700°

As such, the total angular distance reached is 1700°.

Now converting this to radians since both degree and radian outputs are required:

Since 2π equals 360°, it follows that one degree equates to about 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1700° * 2π/360° = 29.7 rad

∅ (rad) = 29.7 rad

For the second part, remember that angular displacement is determined by the angular distance divided by time, leading to a displacement of zero because the knee's ending position is the same as its starting position.

I hope this is helpful!!!!

π

5 0
3 months ago
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