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kherson [118]
2 years ago
13

As an audio CD plays, the frequency at which the disk spins changes.

Physics
1 answer:
Yuliya22 [10]2 years ago
3 0

Answer:

The velocity is 2.5997 m/s.  

Explanation:

The given condition :-

1) N = 210 rpm

speed ( v ) = 1.3 m/s.

We know that ,

v = r ω

where r = radius of disk.

ω = angular velocity.

But ω = \frac{2*\pi*N }{60}

v = \frac{r* 2*\pi*210 }{60}

1.3 = \frac{r*2*\pi*210 }{60}

r = 0.059144 m

2) N = 420 rpm

v = \frac{r* 2*\pi*N }{60} = \frac{0.059144*2*3.14*420}{60}    

v = 2.5997 m/s.  

So we get the velocity v = 2.5997 m/s.  

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The brain receives messages in signals called nerve impulses. Which part of the ear first generates these nerve impulses?
liberstina [14]

That's the job of the tiny "hair cells", located in the <em>inner ear.</em>

If you're a sound wave, this is how you reach the hair cells:

-- go into the big funnel of skin on the outside of the head, that thing we call the "ear"

-- go about an inch or two, down through a skinny dark tunnel inside the skull

-- at the end of the tunnel, hit a dead end, made of a wall of thin skin like a drum, called the "ear drum";  sound waves hit the ear drum and make it vibrate

-- on the other side of the ear drum, inside, is the chamber called the "middle ear".  In there are the three smallest bones in the body; the ear drum touches the first one and makes it vibrate; the first one touches the second one and makes it vibrate; the second one touches the third one and makes it vibrate;  then the third one touches another dead end made of thin skin.

-- the region on the other side of this wall of thin skin is the "inner ear";  it's a long skinny chamber, called the "cochlea",  wound up in a spiral and filled with liquid; the walls of the cochlea are lined with millions of tiny hairs, sticking out into the liquid; the vibrations make waves in the liquid, and the waves make the tiny hairs wave back and forth; each tiny hair is the end of a nerve that goes into the brain; when that hair wiggles, it sends a nerve "message" into the brain.  

-- there are two complete copies of this whole structure ... one on each side of your head.

6 0
2 years ago
Read 2 more answers
A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li
statuscvo [17]

Answer:

37357 sec  

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J  

Energy burns in 1 lift = m g h

                                  = 80 x 9.8 x 1 = 784 J

lifts required = \frac{(14.644 x 10^6)}{784}

                      = 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec  

or 622 min

or 10.4 hrs

3 0
1 year ago
Consider the position vs. time graph below for a woman's movement in a hallway. What is the woman's velocity from 4 to 5 s?
Ksenya-84 [330]

Answer:

The answer is "6\  \frac{m}{s}"

Explanation:

The formula for velocity:

\to \overline{v}={\frac{\Delta x}{\Delta t}}

      =\frac{6}{1}\\\\=6\  \frac{m}{s}

7 0
1 year ago
A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helic
andreev551 [17]

Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s  ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

8 0
2 years ago
for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha
VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

R=\dfrac{u^2\ \sin2\theta}{g}

Here,

u is the speed of launch of projectile

Here, \theta=30^{\circ}

We need to find the other launch angle when the projectile have the same range, such that,

\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}

\sin(60)=\sin2\alpha

\dfrac{\sqrt3}{2}=\sin2\alpha

\alpha =30^{\circ}

So, the other angle is 30 degrees. Hence, this is the required solution.

3 0
2 years ago
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