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rjkz [21]
4 years ago
5

A series of pulses of amplitude 0.25 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? Answer in units of m.
Physics
1 answer:
Klio2033 [76]4 years ago
4 0

Answer:

0m

Explanation:

since the string is rigidly attached to the post, the reflected and incident pulses are of the same amplitude but different polarities. At the point where the two pulses meet, the amplitude will be the addition of that of the incident and the reflected pulses. i.e Amplitude= 0.25 + -0.25=0m

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What current flows through a 15 ohm fixed resistor when it operates on a 120-volt outlet?
gavmur [86]

Answer: current is 8.0 A

Explanation: R= U/I I = U/R = 120 V/15 Ω= 8.0 A

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3 years ago
Un auto se mueve con velocidad constante de 20 m/s y la mantiene durante 20 s. Después se le aplica una aceleración constante y
pentagon [3]

Answer:

26.67 s

Explanation:

v = Velocidad final = 40 m/s

u = Velocidad inicial = 20 m/s

t_1 = Tiempo inicial = 20 s

t_2 = Tiempo empleado durante la aceleración

a = Aceleración

s = Desplazamiento del automóvil durante la aceleración = 200 m

De las ecuaciones lineales de movimiento tenemos

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{40^2-20^2}{2\times 200}\\\Rightarrow a=3\ \text{m/s}^2

v=u+at_2\\\Rightarrow t_2=\dfrac{v-u}{a}\\\Rightarrow t_2=\dfrac{40-20}{3}\\\Rightarrow t_2=6.67\ \text{s}

El tiempo necesario para acelerar es 6.67 s

El tiempo total necesario para toda la ruta es t=t_1+t_2=20+6.67=26.67\ \text{s}.

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3 years ago
Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

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