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telo118 [61]
2 years ago
5

A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflect

ed at the post and travel back along the string without loss of amplitude. What is the net displacement at a point on the string where two pulses are crossing
Physics
1 answer:
AleksAgata [21]2 years ago
8 0

Answer:

 A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

           A_res = 2A

           A_resultant = 2 .01

           A_resulting = 0.2 m

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Wavelength = (wave speed) / (frequency)

Wavelength = (9 m/s) / (0.5 Hz)

<em>Wavelength = 18 m</em>

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The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discover
givi [52]

Answer:

Acceleration due to gravity, a=4.61\times 10^{-14}\ m/s^2

Explanation:

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a=4.61\times 10^{-14}\ m/s^2

So, the acceleration due to gravity at Neptune due to Pluto is 4.61\times 10^{-14}\ m/s^2. Hence, this is the required solution.  

4 0
3 years ago
Look at the picture above and scientifically explain how eating a hamburger can cause energy transformation in three ways within
sergejj [24]

Answer:

CLAIM: When you eat the hamburger, your body uses it and breaks it down which is chemical energy than use that food for energy for mechanical energy which makes thermal energy

EVIDENCE: This happens every day to people across the world and also it is scientifically proven

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Explanation:

7 0
2 years ago
parallel-plate capacitor is made of two square plates 25 cm on a side and 1.0 mmapart. The capacitor is connected to a 50.0-V ba
noname [10]

Answer:

6.9 x 10^-7 J  

3.5 x 10^-7 J

Explanation:

<u>Identify the unknown:  </u>

The energies stored in the capacitor before and after the plates are pulled farther apart  

<u>List the Knowns: </u>

Voltage of the battery: V = 50 V

Area of the plates: A = 0.25 x 0.25 = 0.0625 m^2

Original distance between the plates: d = 1 mm = 10^-3 m

New distance between the plates: d = 2 mm = 2 x 10^-3 m

Permittivity of free space: ∈o = 8.85 x 10^-12 C^2/Nm^2-

<u>Set Up the Problem:   </u>

Capacitance of a parallel-plate capacitor:  

C=∈o*A/d

Energy stored in a capacitor:  

U_c=(1/2)*V^2*C

<u>Solve the Problem:   </u>

<u>Before the plates are pulled farther apart:  </u>

C = 8.85 x 10^-12 x 0.0625/10^-3 = 5•53 x 10^-10 F  

U_c = (1/2) x (50)^2 x 5.53 x 10^-10= 6.9 x 10^-7 J  

<u>After the plates are pulled farther apart:  </u>

C = 8.85 x 10^-12 x 0.0625/2*10^-3 = 2•77 x 10^-10 F  

U_c = (1/2) x (50)^2 x 2•77 x 10^-10 = 3.5 x 10^-7 J  

The energy decrease because the capacitance decrease, so the stored charge decrease and transferred to the battery  

3 0
3 years ago
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