An alarm clock draws 0.5 A of a current when connected to a 120 volt circuit. Calculate its resistance

A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s

sergey [27]

Answer:

Y = V / f where Y equals wavelength

4 Y1 = V / f1 for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2 for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1 dividing equations

f2 = 2 f1

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

8 0

3 years ago

Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32. 4 cm. If he measures the fr

sineoko [7]

The true statement about the wave is that, the wave has traveled 97. 2 cm in 1 second.

In Physics, we define a wave as a disturbance along a medium that transfers energy. The wavelength of a wave is the distance covered by the wave while the frequency of the wave is the number of cycles of the wave completed per second.

The period of the wave is the inverse of the frequency of the wave. It is defined as the time taken for the wave to complete a cycle and it is measured in seconds.

The wave formula is given as;

v = λf

v = velocity of the wave (distance traveled by the wave in one second)

λ = wavelength of the wave

f = frequency of the wave

So;

λ = 32.4 cm

f = 3 hertz

v = 32.4 cm × 3 hertz

v = 97. 2 cms-1

Hence, the true statement about the wave is that, the wave has traveled 97. 2 cm in 1 second.

Learn more: brainly.com/question/14588679

7 0

2 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$