Can objects in a system have momentum while the momentum of the system is zero? explain your answer.

1 answer:

3 0

Yes. Momentum is a vector, so several non-zero momenta can add up to zero. Consider for a simple example a mass of 10kg moving to the right at 5m/s, and another mass of 10kg moving to the left at 5m/s. These momenta each have a magnitude equal to 50 kg m/s but have opposite sign and so the total system's momentum is zero.

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A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s

sergey [27]

Answer:

Y = V / f where Y equals wavelength

4 Y1 = V / f1 for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2 for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1 dividing equations

f2 = 2 f1

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

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when an object falls, eventually the force of air resistance (drag) = the force of gravity. This is called ____

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3 years ago

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Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$