Si units or Systeme' de Internationale' is a widely adopted unit system in measuring basic and derived dimensions In this case, the SI units here are kilograms, meter and seconds. Pounds is an English unit. mass is the measure of <span>how much matter an object contains, hence the answer is A. 43 kg.</span>
Answer:
T = 60 s
Explanation:
There are 6 poles on the track which are equally spaced
so the angular separation between the poles is given as
![\theta = \frac{2\pi}{6}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B6%7D)
![\theta = \frac{\pi}{3}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7B%5Cpi%7D%7B3%7D)
so the angular speed of the train is given as
![\omega = \frac{\theta}{t}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B%5Ctheta%7D%7Bt%7D)
![\omega = \frac{\pi}{30} rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B%5Cpi%7D%7B30%7D%20rad%2Fs)
now we have time period of the train given as
![T = \frac{2\pi}{\omega}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Comega%7D)
![T = \frac{2\pi}{\frac{\pi}{30}}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Cfrac%7B%5Cpi%7D%7B30%7D%7D)
![T = 60 s](https://tex.z-dn.net/?f=T%20%3D%2060%20s)
The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.
From the question given above, the following data were obtained:
Velocity of sound (v) = 343 m/s
Distance (x) = 8.42 m
Time (t) =?
We can obtain obtained the time as illustrated below:
v = 2x / t
343 = 2 × 8.42 / t
343 = 16.84 / t
Cross multiply
343 × t = 16.84
Divide both side by 343
t = 16.84/343
t = 0.05 s
Thus, the time between when the bat emits the sound and when it hears the echo is 0.05 s.
<h3>
How does a bat know how far away something is?</h3>
A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.
Learn more about time elapses between when the bat emits the sound :
<u>brainly.com/question/16931690</u>
#SPJ4
Correction question:
A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)
Answer:
42244138.951 m
Explanation:
G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²
r = Radius of orbit from center of earth
M = Mass of Earth = 5.98 × 10²⁴ kg
m = Mass of Satellite
The satellite revolves around the Earth at a constant speed
Speed = Distance / Time
The distance is the perimeter of the orbit
![v=\frac{2\pi \times r}{24\times 3600}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%5Cpi%20%5Ctimes%20r%7D%7B24%5Ctimes%203600%7D)
The Centripetal force of the satellite is balanced by the universal gravitational force
![m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m](https://tex.z-dn.net/?f=m%5Cfrac%7Bv%5E2%7D%7Br%7D%3D%5Cfrac%7BGMm%7D%7Br%5E2%7D%5C%5C%5CRightarrow%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B2%5Cpi%20%5Ctimes%20r%7D%7B24%5Ctimes%203600%7D%5Cright%29%5E2%7D%7Br%7D%3D%5Cfrac%7B6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%7D%7Br%5E2%7D%5C%5C%5CRightarrow%20%5Cleft%28%5Cfrac%7B2%5Cpi%20%5Ctimes%20r%7D%7B24%5Ctimes%203600%7D%5Cright%29%5E2%3D6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%5C%5C%5CRightarrow%20r%5E3%3D%5Cfrac%7B6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%5Ctimes%20%2824%5Ctimes%203600%29%5E2%7D%7B%282%5Cpi%29%5E2%7D%5C%5C%5CRightarrow%20r%3D%5Cleft%28%5Cfrac%7B6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%5Ctimes%20%2824%5Ctimes%203600%29%5E2%7D%7B%282%5Cpi%29%5E2%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5C%5C%5CRightarrow%20r%3D42244138.951%5C%20m)
The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m
Answer:
Explanation:
70.0(5.00)(9.81) = 3,433.5 = 3430 N