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tigry1 [53]
3 years ago
15

A tube closed at one end is used to determine the speed of sound in air. The resonances occur every 32 cm when a 530-Hz tuning f

ork is vibrated. What is the velocity of the sound?
Physics
1 answer:
soldi70 [24.7K]3 years ago
6 0

Answer:169.6m/s

Explanation:

Wavelength=32cm=0.32m

Frequency=580Hz

Velocity=frequency x wavelength

Velocity=580x0.32

Velocity=169.6m/s

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Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s
AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0
3 years ago
An electric current is flowing through a long cylindrical conductor with radius a. The current density J is uniform in the cylin
RUDIKE [14]

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

4 0
3 years ago
A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio
Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  \frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon  \frac{\omega}{W_n})^2}]

where;

\epsilon = 0

W_n = \sqrt { \frac{k}{m}}

= \sqrt { \frac{100*32.2}{200}}

= 4.0124

replacing them into the above equation and making X the subject of the formula:

X = Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}

X = 0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}

X = 5.676*10^{-3} \ mm

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

8 0
3 years ago
How many significant digits should the answer to the following problem have? (2.49303 g) * (2.59 g) / (7.492 g) =
Feliz [49]

The number of significant digits to the answer of the following problem is four.

<h3>What are the significant digits?</h3>

The number of digits rounded to the approximate integer values are called the significant digits.

The following problem is

(2.49303 g) * (2.59 g) / (7.492 g) =

On solving we get

= 0.86184566204

The answer is approximated to  0.86185

Thus, the significant digits must be four.

Learn more about significant digits.

brainly.com/question/1658998

#SPJ1

6 0
2 years ago
A typical adult can deliver about 12.5 N·m of torque when attempting to open a twist-off cap on a bottle. Assume that bottle cap
Nikitich [7]
Uhhhhhhhhh just tryna get a point so I can ask a question so eh I’m using ur question heheheheheh
3 0
3 years ago
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