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morpeh [17]
3 years ago
14

A series of pulses of amplitude 0.28 m are sent down a string that is attached to a post at one end. The pulses are reflected at

the post and travel back along the string without loss of amplitude. What is the amplitude at a point on the string where two pulses cross if the string is rigidly attached to the post?
Physics
2 answers:
andreyandreev [35.5K]3 years ago
5 0

Answer:

The answer is zero displacement (0 m)

Explanation:

If the end of the string is not free, the reflected pulse has the same amplitude but opposite polarity to the incident pulse. Because of this, for A, the result equals zero (0 m).

irina1246 [14]3 years ago
3 0

Answer:

a. The pulses cancel each other resulting in zero displacement.

b. The pulses reinforce each other, having a displacement of 2× amplitude or 0.56 m.

Explanation:

When the pulse are sent down the attached string  it gets reflected and we have crossing pulses as the incident and reflected pulses cross each other.

a. If the string is rigidly attached to the post then the incident and reflected pulses will have the same amplitude but different direction. That is either the reflected will be going up and the incident down thereby resulting in a cancellation or zero displacement

Incident amplitude = 0.28 m

Reflected amplitude = -0.28 m

Sum = 0.28 m - 0.28 m = 0 m

b. If the end at which the relfection occurs is free to slide, then the incident and the reflected pulses will again have the same amplitude and in this case, the same direction. Therefore;

Incident amplitude = 0.28 m

Reflected amplitude = 0.28 m

Sum = 0.28 m + 0.28 m = 0.56 m

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The eyes of some reptiles are sensitive to 850 nm light. If the minimum energy to trigger the receptor at this wavelength is 3.1
mr_godi [17]

Answer:

Minimum number of photons required is 1.35 x 10⁵

Explanation:

Given:

Wavelength of the light, λ = 850 nm = 850 x 10⁻⁹ m

Energy of one photon is given by the relation :

E=\frac{hc}{\lambda}    ....(1)

Here h is Planck's constant and c is speed of light.

Let N be the minimum number of photons needed for triggering receptor.

Minimum energy required for triggering receptor, E₁ = 3.15 x 10⁻¹⁴ J

According to the problem, energy of N number of photons is equal to the energy required for triggering, that is,

E₁ = N x E

Put equation (1) in the above equation.

E_{1}=N\times\frac{hc}{\lambda}

Substitute 3.15 x 10⁻¹⁴ J for E₁, 850 x 10⁻⁹ m for λ, 6.6 x 10⁻³⁴ J s for h and 3 x 10⁸ m/s for c in the above equation.

3.15\times10^{-14} =N\times\frac{6.6\times10^{-34}\times3\times10^{8}}{850\times10^{-9}}

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8 0
3 years ago
Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?.
natita [175]

For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s   is mathematically given as

F= 618.9 N

<h3>What is the centripetal force?</h3>

Generally, the equation for the angular speed  is mathematically given as

w = v/R

Therefore

w= 4.7/1.8

w= 2.611 rad/s

Where total momentum

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Tm= 915.28

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Ti= 430.24

In conclusion, centripetal force

F= mrw^2

F = m*R*w2^2

F = 76*1.8*2.127^2

F= 618.9 N

Read more about mass

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CQ

Flag

a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?

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Answer:

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