How can a single crash of a car into a wall involve three separate collections?

Two loudspeakers placed 8.0 m apart are driven in phase by an audio oscillator whose frequency range is 2.2 kHz to 2.9 kHz. A po

My name is Ann [436]

Answer:

The answer to the question is 2.2khz

Explanation:

Let z₁ = 5.4m

Let z₂ = 4.6m

The path difference Δz = z₁-z₂ = 5.4 - 4.6 = 0.8m

For the interference= Δz λ, 2λ, 3λ......

The wavelength λ = 0.8m

The speed of sound v = 344m/s

The frequency f = v/λ = 344/0.8 = 430hz

Now,

f₁ =f, f₂= 2f, f₃ = 3f, f₄= 4f, f₅ =5f which is,

f₁ =f = 430Hz, f₂=2f =860Hz, f₃ =3f =1290Hz f₄ =4f =1720Hz and f₅=5f =2150Hz

f5 = 2120Hz = 2.200Hz

we will convert to two significant figures =2.2kHz

8 0

3 years ago

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn

KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.