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Anettt [7]
3 years ago
12

Sound waves travel in air at 343 m/s. The lowest frequency one can hear is 25.0 Hz; the highest frequency is 25.0 kHz. Find the

wavelength of sound for 25.0 Hz and 25.0 kHz
Physics
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer:

13.72 m and 0.01372 m respectively

Explanation:

Wavelength: This can be defined as the distance covered in one complete oscillation. The S.I unit of wavelength is meter (m).

The formula for the speed of a wave is given as

v = λf ............................. Equation 1

Where v = speed of the sound wave, λ = wavelength, f = frequency of the sound wave.

make λ the subject of the equation,

λ = v/f ......................... Equation 2

For the lowest frequency,

Given: f = 25 Hz, v = 343 m/s.

Substitute into equation 2

λ = 343/25

λ = 13.72 m.

For the highest frequency,

Given: f = 25 kHz = 25000 Hz, v = 343 m/s

Substitute into equation 2

λ = 343/25000

λ = 0.01372 m.

The wavelength of sound for 25 Hz and 25 kHz = 13.72 m and 0.01372 m respectively

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sergeinik [125]

Answer:

1500 Joules

Explanation:

Work = Force x Distance

When multiplying by 10 you simply shift all the digits to the

left and append a 0 to the end.

so 150 x 10 = 1500 Joules

3 0
3 years ago
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A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Tems11 [23]

Answer:

a) V_{wf} = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

m_{b}V{b_{i} }  + V_{wi}  = m_{w} V_{wf} + m_{b}V_{bf}

where m_{b},V{b_{i} are the mass and the initial velocity of the bullet, m_{w} and V_{wi} are the mass and the initial velocity of the wooden block, and V_{wf} and V_{bf} are the final velocities of the wooden block and the bullet

The wooden block is initial at rest (V_{wi} = 0) this yields

m_{b}V{b_{i} }  = m_{w} V_{wf} + m_{b}V_{bf}

By solving for V_{wf} adn substitute the givens

V_{wf} = \frac{m_{b}V_{bi} - m_{b} V_{bf}    }{m_{W} }

= \frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}

V_{wf} = 4.67m/s

b) The center of mass speed is defined as

V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}

substituting:

V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)

V = 8.29 m/s

7 0
3 years ago
How does a physicist answer a scientific question?
egoroff_w [7]

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He does an experment.

6 0
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the height of seven falls in colorado is 5/2 the height of twin falls in idahi. the sum of the two heights is 420 ft. find the h
Masja [62]
Let height of twin falls = x
height of seven falls = 2.5x

x + 2.5x = 420
3.5x = 420
x = 420/3.5 = 120

so twin falls = x = 120 ft
seven falls = 2.5x = 300 ft
6 0
3 years ago
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a m
marusya05 [52]

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

4 0
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