P wave for 100km time= distance/speed , T=100km/6.1km/s

A car has two horns, one emitting a frequency of 199 hz and the other emitting a frequency of 203 hz. what beat frequency do the

andriy [413]

Beat frequency, fb = |f2-f1|

That is, beat frequency is the absolute difference between two frequencies. Is is as a results of destructive and constructive inferences.

Therefore, in this case:

fb = 203 - 199 = 4 Hz

3 0

3 years ago

Why was Alessandro Volta's battery important to the evolution of electromagnetics and on the present

Sonbull [250]

Alessandro Volta's battery was a simple and reliable source of electric current, which allowed scientists to study electricity better than they could with previous sources, such as the Leyden jar, and allowed the development of new technology powered by electricity.

3 0

3 years ago

Whenever two apollo astronauts were on the surface of the moon, a third astronaut orbited the moon. assume the orbit to be circu

almond37 [142]

Missing question:
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"

Solution

part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude ( $h=430 km=4.3 \cdot 10^5 m$ ) of the orbit:
$r= r_0 + h=1.7 \cdot 10^6 m + 4.3 \cdot 10^5 m=2.13 \cdot 10^6 m$
This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude, $g=1.08 m/s^2$ . So we can write
$g=a_c= \frac{v^2}{r}$
where $a_c$ is the centripetal acceleration and v is the speed of the astronaut. Re-arranging it we can find v:
$v= \sqrt{g r}= \sqrt{(1.08 m/s^2)(2.13 \cdot 10^6 m)}=1517 m/s = 1.52 km/s$

part b) The orbit has a circumference of $2 \pi r$ , and the astronaut is covering it at a speed equal to v. Therefore, the period of the orbit is
$T= \frac{2 \pi r}{v} = \frac{2\pi (2.13 \cdot 10^6 m)}{1517 m/s} =8818 s = 2.45 h$
So, the period of the orbit is 2.45 hours.

6 0

3 years ago

) a wrench 0.6 meters long lies along the positive y-axis, and grips a bolt at the origin. a force is applied in the direction o

Lera25 [3.4K]

<span>Applied Force Direction vector = [0,1,3] force F = j + 4k torque is t = 100 Newton-meters = r x F Wrench is 0.5 meters long on positive side of the y-axis, r = 0.6 = [0,0.6,0] We know torque |t| = |r x F| = |r| x |F| sin theta r x F = |r| x |F| cos theta r x (j + 3k) = |r| x |j + 3k| cos theta => [0,0.6,0] [0,1,3] = 0.6 x squareroot of ((0)^2 + (1)^2 + (3)^2) cos theta => 0.6 = 0.6 x squareroot of (1 + 9) cos theta => cos theta = 1 / squareroot of (10) Calculationg the sin theta, sin theta = squareroot of (1 - (1 / squareroot of (10))^2) = squareroot of (9/10) sin theta = 3 / squareroot of (10) Substituting the values, |T| = |r| x |F| sin theta => 100 = |0.6| x |F| x 3 / squareroot of (10) |F| = (100 x squareroot of (10)) / 1.8 |F| = (1000 / 18) x squareroot of (10) Magnitude of force |F| = 55.55 x squareroot of (10)</span>

6 0

3 years ago

Which of newton’s laws is connected to observations made by galileo?

Mnenie [13.5K]

Answer:

laws of gravity

hope it helps

5 0

2 years ago