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konstantin123
1 month ago
15

5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be

changed to raise the wave speed to 180 m/s?
Physics
1 answer:
Sav [3.1K]1 month ago
3 0

Answer:

The tension in the string when the speed increased is 134.53 N

Explanation:

Given;

Tension in the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The wave speed is expressed as;

v = \sqrt{\frac{T}{\mu} }

where;

μ represents mass per unit length

v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}

The new tension T₂ will be computed as;

T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N

Consequently, the tension in the string when the speed was increased is 134.53 N

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Ostrovityanka [3204]

Answer:

The beats frequency measures approximately

4.4 kHz

Explanation:

The beat frequency arises from the original ultrasound frequency, f=41.2 kHz, and the frequency of the sound reflected off the car, f':

f_B = f'-f (1)

To calculate the frequency of the reflected sound, we apply the Doppler effect formula:

f'=\frac{v}{v-v_s}f

where

v = 340 m/s, the speed of sound

v_s =33.0 m/sis the velocity of the car

f=41.2 kHzis the frequency of the sound emitted

By substituting values,

f'=\frac{340 m/s}{340 m/s-33.0 m/s}(41.2 kHz)=45.6 kHz

Thus, the beat frequency (1) is

f_B = 45.6 kHz - 41.2 kHz=4.4 kHz

3 0
1 month ago
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A capacitor with initial charge q0 is discharged through a resistor. What multiple of the time constant τ gives the time the cap
Maru [3345]
a) 0.13*τ; b) 2.08*τ. To describe the discharging process of a capacitor through a resistor, consider the following: Q(t) = Qo * exp(-t/τ) to signify a loss of 1/8 of its charge. In this scenario, Q(t) = 7/8 * Qo = 7/8 * exp(-t/τ). By rearranging, we have ln(7/8)*τ = -t, thus t = -ln(7/8)*τ = 0.13. For a loss of 7/8 of its charge, we use Q(t) = 1/7 * Qo * exp(-t/τ), leading to t = -ln(1/8)*τ = 2.08.
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1 month ago
If a person weighs 882 N on the surface of the earth, at what altitude above the earth’s surface must they be for their weight t
Sav [3153]

Para calcular el peso utilizamos la fórmula:
F=m*g=882N

Cuando tenemos dos cuerpos, empleamos la fórmula de la gravedad general:
F=G* \frac{ M_{p}*M_{E} }{ r^{2} }
Donde:
G = constante de gravedad = 6.67* 10^{-11} m^{3} kg^{-1} s^{-2}
Mp = masa de la persona = 882 / 9.81= 89.9kg
ME = masa de la Tierra = 5.97* 10^{24} kg
r = distancia entre la Tierra y la persona

A partir de estas dos fórmulas, deducimos que los lados izquierdos son equivalentes, por lo tanto, los lados derechos también deben serlo.
m*g=G* \frac{ M_{p}*M_{E} }{ r^{2} }

Resolviendo esta ecuación para r:
r= \sqrt{G* \frac{ M_{p}*M_{E}}{M_{p}*g}
r=6371116m = 6371.116km

Esta es la distancia desde el centro de la Tierra. El radio de la Tierra es 6370km y la altura sobre la superficie es 6371.116 - 6370 = 1.116km o 1116m.

3 0
2 months ago
A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108
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Answer: The calorimeter's heat capacity is 6.72J/g^oC

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This scenario assumes the amount of heat lost by the hot object equals the amount of heat gained by the cold object.

q_1=-q_2

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where,

c_1 = specific heat capacity of water = 4.184J/g^oC

c_2 = specific heat capacity of calorimeter =?

m_1 = mass of water = 108.7 g

m_2 = mass of calorimeter = 108.7 g

T_f = final temperature of the mixture = 35.0^oC

T_1 = initial temperature of the water = 60.2^oC

T_2 = initial temperature of calorimeter = 19.3^oC

Now substituting all provided values into the formula, we obtain

(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC

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