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3 years ago

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A guitar string vibrates at a frequency of 330Hz with wavelength 1.40m. The frequency and wavelength of this sound wave in air (

20 C) when it reaches our ears is: Lower frequency, same wavelength Same frequency, same wavelength Higher frequency, same wavelength Same frequency, shorter wavelength

1 answer:

bixtya [17]3 years ago

7 0

Answer:

Same frequency, shorter wavelength

Explanation:

The speed of a wave is given by

$v=f\lambda$

$\lambda=\dfrac{v}{f}$

where,

f = Frequency

$\lambda$ = Wavelength

It can be seen that the wavelength is directly proportional to the velocity.

Here the frequency of the sound does not change.

But the velocity of the sound in air is slower.

Hence, the frequency remains same and the wavelength shortens.

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A 70 kg man's arm, including the hand, can be modeled as BIO 75-cm-long uniform rod with a mass of 3.5 kg. When the man raises b

Butoxors [25]

Answer:

Δy = 7.1 cm

Explanation:

The center of mass of a body is defined

$y_{cm}$ = 1 /M ∑ $m_{i} y_{i}$ i

Where M is the total mass of the body, m mass of each part and ‘y’ height

Let's apply this equation to our case

We locate the reference system on the shoulders

The height of the arms is at its midpoint

y = -75/2 = 37.5 cm

With arms down

$y_{cm}$ = 1/70 (63 y₀ - 3.5 37.5 - 3.5 37.5)

$y_{cm}$ = 1/70 (63 y)₀ - 7 37.5)

With arms up

$y_{cm}$ ’= 1/70 (63 y₀ + 3.5 y + 3.5 y)

$y_{cm}$ ’= 1/70 (63y₀ + 7 35.5)

let's subtract the two equations

$y_{cm}$ ’ - $y_{cm}$ = 1/70 2 (7 35.5)

Δy = $y_{cm}$ ’ - $y_{cm}$ = 2 7 35.5 / 70

ΔY = 7.1 cm

7 0

2 years ago

A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

4 0

2 years ago

When objects are heated they tend to expand because

densk [106]

Hello!
===
When objects are heated, their molecules tend to vibrate fast. As they vibrate, the space between each atom increases. This keeps on happening, and the object expands until it has cooled down.
===
Hope this helps! :)

5 0

3 years ago

Covalent bonds are formed between BLANK

ololo11 [35]

Covalent bonds are formed through an electrostatic attraction between two oppositely charged ions. Hope this Helps :)

6 0

3 years ago

Read 2 more answers

You leave on a 450 miles trip in order to attend a meeting that will start 10.8 hours after you begin your trip. Along the way y

Lorico [155]

Answer:2.6 h

Explanation:

Given

Total Trip distance=450 miles

Meeting starts after 10.8 hours

safe Fastest speed is 55 mi/h

so if he drives all the to the meeting with max speed then it takes $=\frac{450}{55}=8.181 h$

and total allowable time is 10.8

Therefore longest time he can spend over dinner is $10.8-8.181 \approx 2.6 hours$

5 0

2 years ago