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PilotLPTM [1.2K]
3 years ago
15

The water level in a vertical glass tube 1.00 m long can be adjusted to any position in the tube. A tuning fork vibrating at 653

Hz is held just over the open top end of the tube, to set up a standing wave of sound in the air-filled top portion of the tube. (That air-filled top portion acts as a tube with one end closed and the other end open.) Take the speed of sound to be 343 m/s. (a) For how many different positions of the water level will sound from the fork set up resonance in the tube’s air-filled portion
Physics
1 answer:
konstantin123 [22]3 years ago
6 0

Answer:

4

Explanation:

Given that length of tube,L=1 m

Frequency ,f=653 Hz

We know that

V= f λ

4 L = nλ

So we can say that

L_n=\dfrac{nV}{4f}

Where n is the odd number .

Here V=343 m/s

a)

For n=1:

L_1=\dfrac{343}{4\times 653}

L_1=0.13\ m

For n=3:

L_3=\dfrac{3\times 343}{4\times 653}

L_3=0.39

For n=5:

L_5=\dfrac{5\times 343}{4\times 653}

L_5=0.65

For n=7:

L_7=\dfrac{7\times 343}{4\times 653}

L_7=0.91

So total positions will be 4.

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The two parents shown in (Figure 1) exert upward forces
timama [110]

Answer:

W = 83 [N] (weight)

Explanation:

We must clarify that the system is in static equilibrium, in such a way that we must perform a sum of forces on the Y-axis equal to zero, in order to determine the downward force corresponding to the weight of the child.

∑Fy = 0

F_{1}+F_{2}-m*g=0\\18+65=m*g\\m*g = 83 [N]

And the mass can be determined as follows:

m*g=83\\m = 83/9.81\\m=8.46[kg]

5 0
3 years ago
A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1
nasty-shy [4]

(a) 1.72\cdot 10^{-5} \Omega m

The resistance of the rod is given by:

R=\rho \frac{L}{A} (1)

where

\rho is the material resistivity

L = 1.20 m is the length of the rod

A is the cross-sectional area

The radius of the rod is half the diameter: r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m, so the cross-sectional area is

A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

- The voltage at this temperature is V = 15.0 V

- The current at this temperature is I = 18.6 A

So, the resistance is

R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega

And now we can re-arrange the eq.(1) to solve for the resistivity:

\rho=\frac{RA}{L}=\frac{(0.81 \Omega)(2.55\cdot 10^{-5} m^2)}{1.20 m}=1.72\cdot 10^{-5} \Omega m

(b) 8.57\cdot 10^{-4} /{\circ}C

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega

The equation that gives the change in resistance as a function of the temperature is

R(T)=R_0 (1+\alpha(T-T_0))

where

R(T)=0.86 \Omega is the resistance at the new temperature (92.0°C)

R_0=0.81 \Omega is the resistance at the original temperature (20.0°C)

\alpha is the temperature coefficient of resistivity

T=92^{\circ}C

T_0 = 20^{\circ}

Solving the formula for \alpha, we find

\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C

5 0
3 years ago
An object moving with a speed or 67m/s and has kinetic energy of 500 J what is the mass of the object
saveliy_v [14]
To solve this equation, simply plug the values into the equation for calculating kinetic energy.

KE = 1/2mv^2
500 = 1/2(m)(67^2)
500 =2244.5m
m = 500/2244.5 = 0.222 kg.
8 0
3 years ago
A speed-time graph is shown below:
VladimirAG [237]

Answer: 0.5 m/s^{2}

Explanation:

Average acceleration a_{ave} is the variation of velocity \Delta V over a specified period of time \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

\Delta V=V_{f}-V_{o} being V_{o}=0 cm/s the initial velocity and V_{f}=4 cm/s the final velocity  (according to the information given from the described graph)

\Delta t=8 s

Then:

a_{ave}=\frac{4 cm/s -0 cm/s}{8 s}}

a_{ave}=0.5 m/s^{2}

5 0
3 years ago
To win the game, a place kicker must kick a
Dafna11 [192]

Answer:

1.86 m

Explanation:

First, find the time it takes to travel the horizontal distance.  Given:

Δx = 52 m

v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s

a = 0 m/s²

Find: t

Δx = v₀ t + ½ at²

52 m = (22.2 m/s) t + ½ (0 m/s²) t²

t = 2.35 s

Next, find the vertical displacement.  Given:

v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s

a = -9.8 m/s²

t = 2.35 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²

Δy = 4.91 m

The distance between the ball and the crossbar is:

4.91 m − 3.05 m = 1.86 m

5 0
3 years ago
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