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11 days ago

A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope

n at one end, is gradually increased from zero until the first position of resonance is observed at ????=31.2 cm. The wire is 129 cm long and is vibrating in its third harmonic. If the speed of sound in air is 340 m/s, what is the speed of transverse waves in the wire? Assume that the displacement antinode in the air column is exactly at the open end.

Physics

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A 230mg piece of gold is hammered into a sheet measuring 23cm x 17cm. What is the thickness of the sheet in meters? Density of g

kicyunya [3294]

The sheet's thickness is calculated to be t= 0.0003 mm. The mass provided is m = 230 mg, equivalent to 0.23 g. The area of the sheet is A= 23 x 17 cm², totaling A= 391 cm². Given the density of gold is ρ = 19.32 g/cm³, we assume the sheet's thickness is t cm. From the equation Mass = Density x Volume, we know m = ρ A t. Substituting the values results in: 0.23 = 19.32 × 391 × t, leading to t = 0.000030 cm, or equivalently, 0.0003 mm as the final thickness.

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1 month ago

A runner runs 4875 ft in 6.85 minutes. what is the runners average speed in miles per hour?

Keith_Richards [3271]

Calculating the average speed is straightforward by using the formula involving distance and time:

average speed = distance / time

Thus, we have:

average speed = 4875 ft / 6.85 minutes

<span>average speed = 711.68 ft / min</span>

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3 months ago

Read 2 more answers

a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio

kicyunya [3294]

Answer:

a) factor $b=\sqrt{2}$

b) factor $b=\frac{1}{2}$

c) factor $b=1$

d) factor $b=1$

Explanation:

For an oscillating spring-mass system, the time period is expressed as:

$T=\frac{1}{f}$

$T={2\pi} \sqrt{\frac{m}{k} }$

where:

$f=$ represents the frequency of oscillation

$m=$ signifies the mass linked to the spring

$k=$ is the spring's stiffness constant

a) If the mass is doubled:

New mass, $m'=2m$

Thus, the new time period:

$T'=2\pi\sqrt{\frac{m'}{k} }$

$T'=2\pi\sqrt{\frac{2m}{k} }$

$T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }$

$T'=\sqrt{2} \times T$

this leads to factor $b=\sqrt{2}$ as per the question.

b) When the stiffness constant is quadrupled, holding other factors constant:

New stiffness constant, $k'=4k$

Thus, the new time period:

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

this results in factor $b=\frac{1}{2}$ as required.

c) When both mass and stiffness constant are quadrupled:

New stiffness, $k'=4k$

New mass, $m'=4m$

Thus, the new time period:

$T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T$

which leads to factor $b=1$ as stated in the question.

d) If amplitude is quadrupled, the time period remains unaffected because T does not depend on amplitude as demonstrated by the equation.

Thus, factor $b=1$

7 0

2 months ago

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

Softa [3030]

Answer:

a) $\Delta{t} = 5.39s$

b) the distance the motorcycle covers is 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ denote the variables. Next, we analyze the motion equation for the accelerating motorcycle alongside the constant speed of the car:

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ represents the motorcycle's speed at time 2

$v_{c}$ is the steady velocity of the car

$v_{0}$ indicates the initial speeds of both vehicle types at time 1

d signifies the distance separating the car and motorcycle at the initial moment

x is the distance the car travels from time 1 to time 2

Solving the equations provides:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

For the second query, we determine x+d by applying the car’s motion equation to compute x:

$x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933$

3 0

3 months ago