Answer:
(a) ω = 1.57 rad/s
(b) ac = 4.92 m/s²
(c) μs = 0.5
Explanation:
(a)
The angular speed of the merry go-round can be found as follows:
ω = 2πf
where,
ω = angular speed = ?
f = frequency = 0.25 rev/s
Therefore,
ω = (2π)(0.25 rev/s)
<u>ω = 1.57 rad/s
</u>
(b)
The centripetal acceleration can be found as:
ac = v²/R
but,
v = Rω
Therefore,
ac = (Rω)²/R
ac = Rω²
therefore,
ac = (2 m)(1.57 rad/s)²
<u>ac = 4.92 m/s²
</u>
(c)
In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:
Centripetal Force = Frictional Force
m*ac = μs*R = μs*W
m*ac = μs*mg
ac = μs*g
μs = ac/g
μs = (4.92 m/s²)/(9.8 m/s²)
<u>μs = 0.5</u>
Answer:
Nitrogen and oxygen are by far the most common; dry air is composed of about 78% nitrogen (N2) and about 21% oxygen (O2). Argon, carbon dioxide (CO2), and many other gases are also present in much lower amounts; each makes up less than 1% of the atmosphere's mixture of gases.
The answers is
D. The acid creates cracks in the rocks, which
allow air to circulate through the rock,
causing it to weather
Wave speed = (wavelength) x (frequency)
= (45 meters) x (9 per second)
= 405 meters per second .
Answer:
a) f ’’ = f₀
, b) Δf = 2 f₀ 
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
f ’’ = fo
f ’’ = fo 
leave the linear term
f ’’ = f₀ + f₀ 2
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ 