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ahrayia [7]
3 years ago
11

The maximum pressure variations the human ear can withstand are about how much above and below atmospheric pressure?

Physics
1 answer:
Crank3 years ago
4 0

The maximum pressure variations the human ear can withstand  above and below atmospheric pressure is around 30 pa. the normal atmospheric pressure is around 101325 pa. hence the variation in the maximum pressure for human ear is very small as compared to the atmospheric pressure. if the ear is exposed to a pressure greater than this , it can cause permanent damage to the ear.

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A wave has a wavelength of 10mm and a frequency of 5 hz what is the speed?
geniusboy [140]
V=fλ
v=5*0.01
Therefore v=0.05
3 0
3 years ago
Read 2 more answers
A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling
Setler [38]
  • m1=1500kg
  • m_2=3000kg
  • v_1=5m/s
  • v_2=7m/s

Using law of conservation of momentum

\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3

\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3

\\ \sf\Rrightarrow 7500-21000=4500v_3

\\ \sf\Rrightarrow -13500=4500v_3

\\ \sf\Rrightarrow v_3=-3m/s

6 0
2 years ago
A spacecraft is moving past the earth at a constant speed of 0.60 times the speed of light. The astronaut measures the time inte
Afina-wow [57]

Answer:

the time interval that an earth observer measures is 4 seconds

Explanation:

Given the data in the question;

speed of the spacecraft as it moves past the is 0.6 times the speed of light

we know that speed of light c = 3 × 10⁸ m/s

so speed of spacecraft v = 0.6 × c = 0.6c

time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds

Now, from time dilation;

t = Δt₀ / √( 1 - ( v² / c² ) )

t = Δt₀ / √( 1 - ( v/c )² )

we substitute

t = 3.2 / √( 1 - ( 0.6c / c )² )

t = 3.2 / √( 1 - ( 0.6 )² )

t = 3.2 / √( 1 - 0.36 )

t = 3.2 / √0.64

t = 3.2 / 0.8

t = 4 seconds

Therefore, the time interval that an earth observer measures is 4 seconds

6 0
3 years ago
In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the
OlgaM077 [116]

Answer:

a)   m_v = m_s ((\frac{w_o}{w})² - 1) ,  b)  m_v = 1.07 10⁻¹⁴ g

Explanation:

a) The angular velocity of a simple harmonic motion is

           w² = k / m

where k is the spring constant and m is the mass of the oscillator

let's apply this expression to our case,

silicon only

         w₉² = \frac{K}{m_s}

         k = w₀² m_s

silicon with virus

         w² = \frac{k}{m_s + m_v}

          k = w² (m_v + m_s)

in the two expressions the constant k is the same and q as the one property of the silicon bar, let us equal

           w₀²  m_s = w² (m_v + m_s)

           m_v = (\frac{w_o}{w})²  m_s - m_s

           m_v = m_s ((\frac{w_o}{w})² - 1)

b) let's calculate

          m_v = 2.13 10⁻¹⁶ [(\frac{20.4}{2.85})² - 1)]

          m_v = 1.07 10⁻¹⁴ g

4 0
3 years ago
Https://phet.colorado.edu/sims/html/balloons-and-static-electricity/latest/balloons-and-static-electricity_en.html
Katarina [22]

Answer:

there is not pic

Explanation:

5 0
2 years ago
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