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liraira [26]
3 years ago
11

When you voice the vowel sound in "hat," you narrow the opening where your throat opens into the cavity of your mouth so that yo

ur vocal tract appears as two connected tubes. The first is in your throat, closed at the vocal cords and open at the back of the mouth. The second is the mouth itself, open at the lips and closed at the back of the mouth—a different condition than for the throat because of the relatively larger size of the cavity. The corresponding formant frequencies are 800 Hz (for the throat) and 1500 Hz (for the mouth). What are the lengths of these two cavities? Assume a sound speed of 350 m/s.
Physics
1 answer:
xz_007 [3.2K]3 years ago
3 0

Answer: 0.1 m, 0.0583 m

Explanation:

We are given that:

Frequency for throat= 800 Hz

Frequency for mouth= 1500 Hz

Sound speed= 350 m/s

We have to find the corresponding lengths.

We have

f= \frac{v}{4L}

or L=\frac{v}{4f}

For the throat= L= \frac{350}{4*800\\} = 0.1 m

For the mouth= L= \frac{350}{4*1500} = 0.0583 m

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Zigmanuir [339]

Answer:

176.58Watts

Explanation:

Power= work done /time

Where mass(m)=60kg

Height (h) =3m

Time(s)=10s

Force of gravity = 9.81m/s^2

Power=mgh/t

Power= (60kg) * (9.81m/s^2) * (3m)/10s

Power= 176.58Watts

8 0
3 years ago
Starting from rest, a 2.1x10^-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exert
Rainbow [258]

Answer:

a) The flea's speed when it leaves the ground is v=1.88m/s

b) The flea move 18cm upward while it is pushing off

Explanation:

Hi

<u>Knwons</u>

Mass m=2.1\times 10^{-4} kg, Work W=3.7\times 10^{-4} J and Force F=0.41N

a) Here we are going to use W=\frac{1}{2} mv^{2}, so v=\sqrt{\frac{2W}{m} }= \sqrt{\frac{2(3.7\times 10^{-4} J)}{2.1\times 10^{-4} kg} }=1.88m/s

a) Here we are going to use W=mgh, so h=\frac{W}{mg}= \frac{3.7\times 10^{-4} J}{(2.1\times 10^{-4} kg)(9.8m/s^{2})} =0,1797m or 18cm approx.

7 0
3 years ago
Gravity is not considered matter. <br>A. True <br>B. False​
Strike441 [17]

Answer:

false gravity is not considered matter

7 0
2 years ago
A car moving with a speed of 35 m/s sees a child standing in the
zlopas [31]

Answer:

2100 N

Explanation:

v = u + at \\ 0 = 35 + a(5) \\ a =  - 7m {s}^{ - 2}  \\ \\ net \: force \:  = ma \\  = (300)(7) \\  = 2100 \: newtons

6 0
2 years ago
Running at 1.55 m/s, Bruce, the 40.0 kg quarterback, collides with Biff, the 90.0 kg tackle, who is traveling at 7.0 m/s in the
Margarita [4]

Answer:Bruce is knocked backwards at  

14

m

s

.

Explanation:

This is a problem of momentum (

→

p

) conservation, where

→

p

=

m

→

v

and because momentum is always conserved, in a collision:

→

p

f

=

→

p

i

We are given that  

m

1

=

45

k

g

,  

v

1

=

2

m

s

,  

m

2

=

90

k

g

, and  

v

2

=

7

m

s

The momentum of Bruce (

m

1

) before the collision is given by

→

p

1

=

m

1

v

1

→

p

1

=

(

45

k

g

)

(

2

m

s

)

→

p

1

=

90

k

g

m

s

Similarly, the momentum of Biff (

m

2

) before the collision is given by

→

p

2

=

(

90

k

g

)

(

7

m

s

)

=

630

k

g

m

s

The total linear momentum before the collision is the sum of the momentums of each of the football players.

→

P

=

→

p

t

o

t

=

∑

→

p

→

P

i

=

→

p

1

+

→

p

2

→

P

i

=

90

k

g

m

s

+

630

k

g

m

s

=

720

k

g

m

s

Because momentum is conserved, we know that given a momentum of  

720

k

g

m

s

before the collision, the momentum after the collision will also be  

720

k

g

m

s

. We are given the final velocity of Biff (

v

2

=

1

m

s

) and asked to find the final velocity of Bruce.

→

P

f

=

→

p

1

f

+

→

p

2

f

→

P

f

=

m

1

v

1

f

+

m

2

v

2

f

Solve for  

v

1

:

v

1

f

=

→

P

f

−

m

2

v

2

f

m

1

Using our known values:

v

1

f

=

720

k

g

m

s

−

(

90

k

g

)

(

1

m

s

)

45

k

g

v

1

f

=

14

m

s

∴

Bruce is knocked backwards at  

14

m

s

.

Explanation:

5 0
2 years ago
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